THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES "his b ;he SOUTHERN BRANCH, UNIVERSITY OF CALIFORNIA, LIBRARY, LOS ANGELES, GALJF. THE DIRECTIONAL CALCULUS, BASED UPON THE METHODS OF HERMANN GRASSMANN. BY E. W. HYDE, PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF CINCINNATI, CINCINNATI, OHIO. 48807 BOSTON, U.S.A.: PUBLISHED BY GINN & COMPANY. 1890, COPYRIGHT, 1890, BY E. W. HYDE. ALL RIGHTS RESERVED. TYPOGRAPHY BY J. 8. CUSHING & Co., BOSTON, U.S.A. PHESSWORK BY GINN & Co., BOSTON, U.S.A. Engineering & Mathematical Sciences Library QA a THE 4, Mechanics, should be generally introduced to the knowledge of the coming generation of English-speaking mathematicians, PREFACE. wonderful and comprehensive system of Multiple Algebra invented by Hermann Grassmann, and called ^ by him the Ausdehnungslehre or Theory of Extension, though long neglected by the mathematicians even of Germany, is at the present time coming to be more and more appreciated and studied. In order that this system, with its intrinsic natural- ness, and adaptability to all the purposes of Geometry and 1 V it is very necessary that a text-book should be provided, suitable for use in colleges and universities, through which students may become acquainted with the principles of the ^V subject and its applications. The following pages present, in part, the results of eight or nine years of study and experience in lecturing to university classes upon this subject, and the Author hopes that they may aid in some measure to bring about that general study and use of directional methods, which, in his opinion, will ultimately cause the comparatively awkward and roundabout methods of Cartesian coordinates to be superseded by them, for many of the purposes of analysis. As the great generality of Grassrnann's processes all results being obtained for n-dimensional space has been one of the main hindrances to the general cultivation of his iv PREFACE. system, it has been thought best to restrict the discussion to space of two and three dimensions. The Author, though formerly an enthusiastic admirer of Hamilton's Quaternions, has been brought, by study and expe- rience in teaching both, to a firm belief in the great practical, as well as theoretical, superiority of Grassmann's system. This superiority consists, according to the judgment of the writer, first, and largely, in the fact that Grassmann's system is founded upon, and absolutely consistent with, the idea of geometric dimensions. Second, in the fact that all geometric quantities appear as independent units, viz. : the point ; the point at oo , or line direction ; the definite line ; the line at oo , or plane direction ; the definite plane ; and, finally, the plane at oo, equivalent to a volume, which is scalar. The same holds for space of any number of dimensions ; in fact, it seems scarcely possible that any method can ever be devised, compar- able with this, for investigating w-dimensional space. Now Quaternions deals only with the vector, or line direc- tion, and the scalar for a quaternion is only the sum of these two; it knows nothing of a vector having a definite position, which is the complete representative of the space qualities of a force. Further, Hamilton's vector is not a vec- tor pure and simple, but a versor-vector a fact which gives its peculiar character to his system, as being really a calculus of directed imaginaries. This, which Tait regards as " one of the main elements of the singular simplicity of the Quaternion Calculus," appears to the writer in a very different light. It gives rise to such equations as * = ji, i.e. multiplying two quantities together is the same as dividing one by the other, which the author has found a great stumbling-block to the student. It can hardly be regarded as a natural geometrical PREFACE. V conception, that the product of two vectors should be another vector perpendicular to each of them ; still less, that it should be, as in the general case, a scalar plus such a vector. In fact, the Quaternion system practically throws overboard the idea of geometric dimensions. It has been deemed advantageous, however, to make use of certain terms and symbols introduced by Hamilton, such as scalar, tensor, with its symbol T, etc. Although this work is based upon the principles and meth- ods of Grassmann, yet much matter will be found in it that is believed to be original with the author. Thus the idea of the complement in a point system, with its geometric interpretation, as developed in Arts. 40-44 and 63, does not occur in Grassuianu's works, either of 1844 or 1862. This idea is of great value in geometric applications, and cor- responds to that of duality in Modern Geometry. In Arts. 68 and 69 are treated some properties of what I have called screivs, which, as such, are not discussed by Grass- mann. One of the most beautiful and valuable theories developed by Hamilton in his great works is that of linear and vector functions. This theory I have found to be equally capable of application to point functions in. n-dimensional space. In Chapters IV. and VI. this linear, point function has been used in the discussion of the conic and quadric, giving rise to a treatment of these loci bearing the same analogy to trilinear and quadriplanar methods that the vector treatment of these curves and surfaces bears to ordinary Cartesian methods, and possessing likewise the same superiority in clearness, concise- ness, and intelligibility which the vector methods have over the Cartesian. VI PREFACE. It appears to the writer that Hamilton's method of dealing with linear functions much excels that of Grassmann, as devel- oped in the second part of his work of 1862, for practical util- ity and convenience. It is hoped and believed that the exposition of the funda- mental ideas and principles of the system in the first two chapters will be found sufficiently full and explicit to give the student a working knowledge and grasp of the subject, such as will enable him to take up without difficulty the work of the succeeding chapters, or to read, with comparative ease, the original treatises of Grassmann. A large number of exercises have been inserted, in the belief that only by the repeated application of the principles which he has learned to the solu- tion of actual problems can the student acquire any real com- mand of any branch of Mathematics. Eight or nine blank pages will be found at the end of each chapter, for the recep- tion of notes, solutions, etc., the Author's experience having convinced him of the advantages of such an arrangement. E. W. HYDE. TABLE OF CONTENTS. CHAPTER I. ADDITION AND SUBTRACTION. PAGE Definition of vector, plane- vector, etc 1 Equality 1 The point, equality of points 2 Difference of unit points 2 Sum and difference of vectors 3 Addition of points 4 Case in which 2m = 5 Conditions that two vectors may be parallel, and that three may be parallel to one plane 7 Dependence and independence 8 Tensor, unit, reference systems 9 Exercises 10 CHAPTER II. MULTIPLICATION. Fundamental notion of a geometric product 23 Posited quantities 23 Definitions of combinatory products 24 Laws of combinatory multiplication 25 Product of two points 25 Product of two vectors 20 Product of three points 27 Product of three vectors 28 Product of four, or more, points, or vectors 29 Equations of condition, planimetric products 30 Product of points expressed in terms of reference points 31 Planimetric product of two lines 32 Product of a point and two lines 33 Viii TABLE OF CONTENTS. PAGE Conditions L^ = and L^L.^ = 34 Product of two || lines and of two vectors 34 Sum of point- vectors 35 Complement 36 Co-product and co-square 37 Complement in a plane point system 39 Geometric interpretation of the same 40 Distance between two points 41 Anti-polar property of complement 41 Projections 43 Multiplication table for a point system 44 Division indeterminate 49 Exercises 50 Stereometric products 56 Products of points expressed in terms of reference points 56 Product of a line and a plane 57 Product of two planes; of three planes 58 Product of four planes; products of plane-vectors 59 Equations of condition 60 Addition of planes and plane- vectors 60 Addition of point-vectors, or lines 62 The complement in three-dimensional space 62 Complement in a point system in three-dimensional space 64 Condition that a linear function of the edges of the reference tetrac- dron shall be a point-vector 65 Point system in solid space a fifteen- fold algebra 66 Projections in a vector system 67 Projections in a point system 68 Normal form of the screw 71 Product of two screws 72 Products Sfi Sp 2 and Sp l Sp t Sp 3 73 Quaternion expressions corresponding to some expressions of this calculus 73 Exercises . . 74 CHAPTER III. APPLICATIONS TO PLANE GEOMETRY. Preliminary statements 87 Equations of plane space 88 Equations of right lines 89 TABLE OF CONTENTS. IX PAGE Transformation from a point to a vector system, and vice verscl 90 Exercises 91 Equations L 1 p L. 2 p = C and p 1 L p t L = C 92 Conies through three points, and touching three lines 93 Differentiation 94 Examples of differentiation 96 Tangent and normal. The circle 97 Exercises on the circle 98 The parabola 99 Exercises on the parabola 101 The ellipse and hyperbola 102 Tangent and normal. Diameter 104 The

function 114 The invariants m a,nd m, 116 Exercises on inversion 117 The general equation of the second degree in plane space 117 Center of the locus 118 Case when tj>\p shall be the product of two linear factors 140 Nature of the locus at oo 140 Most general form of p\p 141 Conic through the reference points 141 Circle through the reference points 142 Exercises 14:> Equations p\ 150 Exercises 151 CHAPTER V. SOLID GEOMETRY. Preliminary statements 163 Non-scalar point equations of surfaces and curves in general 164 Non-scalar plane equations of surfaces in general 164 Non-scalar vector equations in general 165 Equations of planes, lines and points 165 Vector equations of planes and lines 166 Exercises 167 The sphere 170 Exercises on the sphere 171 The paraboloids 172 Rectilinear generators 173 Exercises on the paraboloids 173 The central quadrics 174 Tangent plane and normal 176 Diametral plane and conjugate diameters 177 Poles and polars 178 Volume of circumscribed parallelepiped 179 Surface referred to conjugate diameters 180 Rectilinear generators 180 Exercises 181 Condition that p\p shall be factorable 183 TABLE OF CONTENTS. XI PAGE The function in general. Inversion 183 The general scalar equation of the second degree in terms of vectors. 184 Cyclic sections 185 Exercises 186 Center of the general quadric surface 188 Maximum and minimum values of Tp 190 Roots of the discriminating cubic 192 The discriminant m C' 194 Table of classification of the quadric 1 95 Exercises . . . 195 CHAPTER VI. SCALAR POINT EQUATIONS OP THE SECOND DEGREE IN SOLID SPACE. Differentiation. The general homogeneous equation 207 Diametral planes. Significance of the quantity \p 212 Rectilinear generators 212 The discriminant 213 Nature of the surface at infinity 214 Table of classification 216 Quadric through the four reference points 217 Conditions that (496) shall represent a sphere 217 Exercises 218 Inversion of

i to p 2 , we write P-Pi = e, ......... (1) and the development of the subject will show this to be in accordance with the above proviso. CHAP. I.] ADDITION AND SUBTRACTION. 3 If p 2 pi = e = 0, or p 2 = PI, there is no difference of position between the points ; i.e. they coincide, as previously stated. Adding p l to both sides of (1), we have in which, of course, the meaning of the sign + is extended in a manner similar to that of the minus sign. Eq. (2) shows that the sum of a point and a vector is a point, distant from the first point by the length of the vector, and in the direction of the same. 6. Let e,, e 2 , e s be any three unit points, and e 1? e,, c 3 vectors || and equal in length to the sides of the triangle e-f,^ taken in order, so that cj = e 3 e 2 , c 2 = &i e s , e 3 = e 2 e t : then that is, the sum of the vector sides of a triangle, taken posi- tively in the same sense around the triangle, is zero. We have also so that the sum of two vectors is a vector which represents the differ- ence of the initial and final posi- tions of a point which moves along the two vectors successively. Write e/ = e 2 - e s = - Cl ; then / 2 = e 2 e. A Ci-\-e 3 = e 2 ei = e s j . . . (4 a) that is, the difference of two vectors is the vector joining their extremities when they are drawn outwards from a common point. Similarly, if we take n unit points e 1} e 2 , e n , we have - i(pi e) + m 2 (p 2 -e) = (m,i + m 2 ) (p e), 1 1 vi / \ , 7/fco / \ / ~ \ or p- e = - -( Pl - e ) + - _(#,_), . . (7) m l + m 2 mi + ^2 by which p is easily found. For three points we have m 1 pi + m 2 p 2 + m s p s = (m 1 -\-m 2 -\-m 3 )(p)', . . (8) or, by (5), (?n,! -f- m 2 )p + m s p 3 = (w^ + m. t + m z } (p) ' ; CIIAF. I.] ADDITION AND SUBTRACTION. 5 so that (j>) ' is on the line joining p and p s at distances from them inversely proportional to their weights. Subtracting from both sides of (8) (m 1 + m 2 + m 3 )e, we have, on dividing by 2w, >-v, 1l - 2 -e) + m s (p s - e) (P) 6 = - - - by which (p) ' is easily constructed. Similarly, for any number of points we have p"2m, ..... ........ (10) and p e = 2[m(p e)] ......... (11) 2m The reader will notice the analogy to "center of parallel forces." 8. Case in which 2m = 0. In (5) and (6) let m l + ni 2 = 0, id these equations become aid jp Pa=.p Pi ............. (13) If m 2 be not zero, and jp a not coincident with p 2) these equations can only be satisfied when p is at an infinite distance on the line pip 2 - Eq. (6) shows at once that, when m a is negative, p is not between pi and p 2 , and that, as the numerical value of m 2 approaches that of m^ p recedes farther and farther, until, when mi + m 2 = 0, j? is at GO. Thus the meaning of eq. (12) is, that a point of zero weight at an infinite distance is equivalent to a vector of definite length directed towards this point. It will often be convenient to regard vectors as points at QO. Consider next the general case of n points. Eqs. (10) and (11) become, when 2^ = 0, and 2"[m(p e)] =0 (p e). 6 DIRECTIONAL CALCULUS. [ART. 8. Let (pY^in = m'(p)', say: then, since mj + -m' = 2"?n = 0, +m'(p)'= m 1 ((p)' -p v ) = 0-p, and 2m-e=m 1 p 1 -e)+ Hence the direction of the point at co is found by constructing the mean point of all the points except one, when the vector from the excepted point to this mean point has the required direction. As p is a unique point for any given system of weighted points, and p t may be taken as any point of the system, it follows that, in any system of points whose total weight is zero, a line drawn from any point of the system to the mean point of all the rest is parallel to any other such line. Finally, suppose 2mp=p-2m = 0, ...... (16) which gives as a necessary consequence 2m = 0. Then, by (15), if m' is not zero, we must have that is, P! is the mean point of the system consisting of the remaining points. But, as p l may be any point of the system, it appears that any one of the points is, in this case, the mean of all the rest. For example, let n = 3 ; then (16) becomes m iPi + W 2 p 2 + m s p s 0, which requires also m t + m 2 -f- m 3 = 0; whence ~ m s p s = (?rcj or p s is the mean point of m^ and m 2 p 2 . Now we have seen in Art. 7 that the mean of two points is collinear with them ; hence Q ...... (17) is the condition that the three points shall be collinear. CHAP. L] ADDITION AND SUBTRACTION. 7 If three points p l} p z , p 3 are not collinear, and yet are con- nected by a linear relation such as (17), this equation can only be satisfied when each iveight is separately zero, that is mi = m 2 = m 3 = ; ........ (17 a) for, as we have just seen, for all values of the weights different from zero the three points are collinear, but when each weight is zero, (17) is satisfied independently of the positions of the points. % Similarly, m\p\ + m 2 p 2 + m s p 3 + m 4 ^4 = 0, ... (18) which requires also m l + m z + m z + m^ = 0, is the condition that four points shall be coplanar, and if the points are not in one plane, then we must have M! = m 2 = m 3 = m 4 = ....... (19) A similar condition between Jive points would imply that the points were all in one tri-dimensional space, which leads to the consideration of space of higher dimensions than three. 9. Let cj and e 2 be two vectors, and 7^ and n 2 scalars ; then e 2 =0 ......... (20) is the condition that the vectors shall be parallel, as appears at once by Art. 2, if we write the equation n^ = Similarly, n^ + w 2 e 2 -f n s e s = ....... (21) requires the three vectors to be parallel to one plane; for, by Art. 6, the vectors n^ n 2 c 2 , ri 3 e 3 , may be represented by the three sides of a triangle. The same appears from (17) if we take pu p 2 , p s as points at GO. If we have the additional condition (22) the extremities of the vectors, if drawn outwards from a point will be in one right line. For let e , e^ e 2 , e s be four points 8 DIRECTIONAL CALCULUS. [Aux. 10. so taken that e 1 e e 1 , ee =c.,, e 3 e = e 3 ; then (21) becomes rti(gj e ) + n 2 (e, e fl ) + n s (e 8 e ) = 0, or, by (22), n^ -+- n 2 e 2 + n 3 e 3 = ; which, by (17), requires e { , e 2 , e s to be collinear. Similarly, it may be shown that the eqs. = sn 4 4 = -j ?i 4 = I are the conditions that the extremities of the vectors tj, e 2 , etc., drawn outwards from one point shall be coplanar. 10. If the equation Swip = can be satisfied by finite values of the ra's different from zero, then, as we have seen, any one of the points can be expressed in terms of the others, and is, therefore, dependent on them ; thus the points are mutually dependent. If, however, the equation can only be satisfied when all the m's are zero, no relation of this kind exists be- tween the points ; i.e. they are independent. This is evidently as true of vectors as of points. Thus, by Art. 4, two different points are always independent. Three non-collinear points are independent, while three col- linear points are mutually dependent. Four non-coplanar points are independent, while four coplanar points are mutually dependent. Any five, or more, points are always mutually dependent. Similarly, two non-parallel vectors are independent, but two parallel vectors are dependent. Three vectors not parallel to one plane are independent, but when all are parallel to one plane they are mutually dependent. Four, or more, vectors are always mutually dependent. In a system of independent points or vectors no one can be expressed linearly in terms of the others. CHAT. I.] ADDITION AND SUBTRACTION. 9 11. The tensor of a directed quantity is its numerical magni- tude, taken always as a positive quantity. It will be denoted, as in Quaternions, by T ; as, Te. = tensor of e. That portion of a directed quantity whose magnitude is unity will be called its unit, and will be denoted by U; as, Ue = unit of e. Hence we have Te Ue = e . (24) 12. Reference systems. Let p be any point, and e , e 1} e 2 three fixed reference points ; then writing p = xe + ye! + ze 2 , (25) we must have, if p is to be a unit point, x + y + z=l, (2G) and p is the mean point of xe , ye and ze 2 . Eliminating x between (25) and (26), we have P = e + y(ei-e l) )-\-z(e 2 -e ) ) .... (27) from which it appears that, by varying y and z, p may be made to occupy any position whatever in the plane of e , e,, e 2 . Hence any 1 three unit points e , e 1? e 2 may be taken as a refer- ence system for plane space, in terms of which all poifTts of this plane space may be expressed. Writing p e fl = p, e, e = e l5 e 2 e = e 2 , eq. (27) becomes P = yi + e (28) and any vector in plane space may be expressed in terms of two reference vectors c^ e 2 in that plane. When ej and e 2 are of unit length and at right angles, the system will be called a unit normal reference system, and ^ and i 2 will be substituted in this case for cj and e 2 . Similarly, in solid space any point p may be expressed in terms of four non-coplanar points e , e 1} e 2 > ^ by the equation p = we + xe l + ye 2 + ze 3 , (29) 10 DIRECTIONAL CALCULUS. [ART. 13. Avhich requires, if p is a unit point, w + + y + z = l, ......... (30) whence p = e -\-x(e l - 3- When Tcj = TV, = 2Vs = 1> and the three vec- tors are at right angles to each other, we have a unit normal system, and, in this case, substitute t 1? t 2 , i 3 for e l9 e 2 , c 3 . 13. A number of exercises will now be given illustrative of the application of the preceding principles. (1) The mean point of unit points at the vertices of a tri- angle coincides with the center of gravity of its area. The center of gravity of the area will be at the common point of lines through the vertices and the middle points of the opposite sides. Thus, if p be the point, 2 (e 2 +e 3 ) =x'e. 2 .-. (x - iy' by eqs. (17) and (17 a) this gives * - iy' = %y x< = y y' = Q. But, since p is a unit point, we have x + y = 1 = x' + y'. From these equations we find x = x' = i, 2/ = y = |; whence p = $ e l + |p, = 1 (e l + e 2 -f e 3 ), CHAP. I.] ADDITION AND SUBTRACTION. 11 which shows that p is the mean point of e\, e 2 , e s , and trisects the distance from e l to p^. (2) To find the common point of the perpendiculars from the vertices of a triangle on the opposite sides. Let Z, m, n be the ratios of the sides of the triangle to the cosines of the opposite angles ; i.e. m/ _ ~ \ -, etc. ; cos(Z at m0 -I- 77.1 then m + 1 _ ne 3 + fcj _ Ze t + me., n-\-l I -+- 7)1 Proceeding now precisely as in the first problem, with the above values of p^ p 2 , p s instead of those there used, we find , _le L + me, 4- ne x l + m + n (3) Find the common point of perpendiculars to the sides of a triangle through the middle points of the respective sides. By the figure for the last problem we have i" Ifo _1_ P" = S(e l + l + m + n m + n Hence, equating to zero coefficients of e l and e 2 , y (m + 71) _ __ 1 a l + m + n I + m + n 2 | 2 .: x=y=-, and p" = 12 DIRECTION, AL CALCULUS. [ART. 13. (4) Show that p, p', p" of examples 1, 2, 3 are collinear. It is easily seen that 3p p' 2p" = 0, which, by (17), proves the collinearity of the points. It is evident also that p, p', and p" are collinear whatever values be assigned to I, ra, n ; for nothing in the demonstration depends on I, m, n having the values assigned in ex. 2. (5) The center of gravity of the volume of a tetraedron coincides with (a) The mean of unit points at the vertices ; (6) The center of gravity of the tetraedron whose vertices are the centers of gravity of its faces ; (c) The mean of the middle points of its edges. (G) Find a point such that the sum of the vectors drawn from it to n given points shall be zero. Ans. p = -2te = mean of points. n (7) Construct by eq. (11) the mean point of points at the four corners of a square, whose weights are respectively 1, 2, 3, and 4. (8) Let e 1} e 2 , e 3 , e 4 , be the corners of a parallelogram, and e & = %(d + e 2 ) : show that e^ and e 4 e 5 will trisect each other. (9) Let 6j, e 2 , e s , e 4 be the corners of a parallelogram, and let p l} p 2 be points on a line || to e^ : if q is the common point of P& and p 2 2, and q 2 the common point of p^ and p 2 e s , show that g,g 2 is || to e^. (10) Let 1, e 2 , e s , e 4 be four non-coplanar points, and let Pi, Pv Ps> P* be points taken on e^, e 2 e 3 , e^ and e i e l respec- tively: find the condition to be fulfilled -in order that p^p 2 , 6163, and p s p 4 may have a common point. Write p l ra^ + w,e 2 , p 2 = m z e 2 -\- n 2 e s , etc. ; then the p's must be coplanar, which leads to the condition CHAP. I.] ADDITION AND SUBTRACTION. 13 (11) If through any point within the triangle e^e-fa lines be drawn || to the sides and terminated by them, and if I, m, n, be the respective ratios of these lines to the sides to which they are || ; then I + m + n = 2. (12) The center of gravity of the sides of a triangle coin- cides with the center of the circle inscribed in the triangle formed by joining the middle points of the sides. (13) Find the center of gravity of the faces of a tetraedron ; also of the edges. CHAP. II.] MULTIPLICATION. 23 CHAPTER II. MULTIPLICATION. 14. Grassmann's first conception of a geometrical product is that it is what is produced or generated by the first factor, as it moves over a distance determined by the second. Thus ejea, if ci and t% are two vectors, signifies the directed plane area bounded by the parallelogram whose sides are \\ and equal in length to cj and 2 ; that is, the plane area gener- ated by EJ as it moves, || to itself, along e 2 from its initial to its final point. EI This is a plane-vector as defined in Art. 1. The product will evidently have the same value whether the initial point of c x moves in the direction e 2 or in any other path, provided that tj itself moves in the plane of cj and e 2 from the initial to the terminal point of e 2 . Evidently e 2 j, interpreted in the same way, gives a genera- tion of the same parallelogram in the reverse sense, and should therefore be the negative of e^ Similarly, if p l and p 2 are two unit points, p t p 2 is that which is generated by pi in moving from its position to that of p 2 in a right line ; thus prf 2 is a line of definite magnitude, direc- tion, and position ; i.e. a point-vector, according to Art. 1. Evidently p 2 pi = Pip* Such products as the above are called combinatory, because the factors combine to form a new geometric quantity different from either of the component factors. 15. The term posited will be applied to such geometric quantities as have definite positions ; as, for instance, a right line or plane passing through a definite point not at oc. 24 DIRECTIONAL CALCULUS. [ART. 16. Any two geometric quantities which differ only in magnitude are said to be congruent. 16. Another conception of combiuatory products of posited quantities will now be given, which will be found to be con- sistent with that given in Art. 14. (a) The product of two posited quantities which have no com- mon figure is some multiple of the connecting figure. (b) TJie product of two posited quantities ichich MUST have a common figure in the space under consideration, is the common figure, multiplied by a scalar quantity. EXAMPLES. Under (a), the product of two points is the connecting straight line as in Art. 14. Under (6), the product of two point-plane-vectors, which must necessarily have a common line, is that common line (point-vector) multiplied by a scalar to be determined here- after. (c) A continued product of several posited quantities is to be interpreted by taking together first the two factors on the right, then the result of this multiplication with the next factor towards the left, then this result ivith the next, etc. ; but if at any stage of the process the product of the factors treated up to this point becomes a scalar quantity, then this scalar will not form a COM- BINATORY product with the next factor to the left, but is to be treated like any other merely numerical factor, obeying as a WHOLE the laws of ordinary algebraic multiplication. This statement cannot well be illustrated here, but its mean- ing will appear in the sequel. Grassmann calls a product of the kind (a) progressive, be- cause it is of a geometric order higher than that of either factor ; while one of the kind (6) is regressive, because it is of lower order than either factor. If in a continued product of factors, as in (c) , some of the successive products are pro- gressive and some regressive, then the product as a whole is said to be mixed. CHAP. II.] MULTIPLICATION. 25 17. Lairs of combinatory multiplication of any number of points or vectors not exceeding the number of INDEPENDENT points or vectors possible in the space under consideration. These are : The associative law ; PlP*Ps=P\'P*P* = PlP*'P* (33) The distributive law ; Pi(pt+Ps)=PiPi+PiPs (34) The alternative law ; PiP 2 = P 2 Pi (35) If p. 2 = p 1 in (35), we have pip\ = Q, which agrees with the meaning of the product of two points as given in Art. 14. Also pMPi = Pi p 2 pi = -Pi- PiP 2 = - PiPi P 2 = 0, \)j (33) and (35) ; hence, if there occur tico identical factors in a product of the kind stated at the head of this article, the product is zero. The meaning of certain products will be different according as they are interpreted in two- or three-dimensional space ; hence we shall apply the term planimetric to such a product when it is to be taken in two-dimensional space, and the term stereometric when it is to be taken in three-dimensional space. We proceed now to a detailed discussion of all the geometric products possible in two- and three-dimensional space. 18. Product of tico points. Let p l and p 2 be two unit points. Their product, Pip^ has already, in Art. 14, been stated to be the portion of the right line Pl 'ft fixed by p l and p 2 extending from PI to p 2 , and it will be called a point-vector, or simply a line for brevity. Let f=p. 2 p 1 ; then Pi(P*-Pi)=Pi* = PiPs-PiPi=PiP - (36) by the last article ; so that the product of two points is equiva- lent to that of the first point into the vector from the first to 26 DIRECTIONAL CALCULUS. [ART. 19. the second. Hence multiplying a vector by a point changes it into a point- vector ; i.e. fixes its position. Let e' be any vector whatever; then, by eq. (2), p l + xe' may be any point in space by suitably choosing x and e : multiply into e ; therefore Hence the point-vector obtained by multiplying the vector e by the point p t + xe' is not in general equal to pje, but differs from it by the quantity xe'e. If, however, we have e' = e, then p l ' i ..... (38) so that, when c is multiplied by any point on the line through Pi and p 2 , the resulting point-vector is equal to p^. Hence, in order that two point-vectors may be equal, they must have the same length, the same direction, and must be situated upon the same straight line, while their position on this line is indifferent. 19. Product of two vectors. Let ^ and e 2 be any two vec- tors : their product e^ has already, in Art. 14, been stated to be a plane area || to ej and e 2 and equal to the area of the par- allelogram whose sides are parallel and equal in length to c t and e 2 . Write and c' = then c and e' may be any two vectors || to the plane-vector e^ by giving suitable values to 1} x 2 , y lf y 2 . Multiplying, we have * (39) Hence the product of any two vectors || to Cjc 2 only differs from cjCjj by a scalar factor. If x$ 2 xfli = 1, then te' = e^. Thus in order that two plane-vectors should be equal, it is only necessary that their plane directions and areas should be the same, without regard to the directions of the component vectors. Of course direction as applied to a plane-vector must CHAP. IT.] MULTIPLICATION. 27 include the sense in which the generation takes place. See Art. 14. Regarding ej and c 2 as points at oc, we see that, just as a zero point at oc is a vector, so the product of two such, that is a point-vector at oo, is a plane-vector ; or, in other words, just as a point at oo gives a line direction, so a line at oo gives a plane direction. 20. Product of three points. Let p t , p% p 3 be three unit points. By Art. 16, (a) and (c), the product should be a multiple of the connecting tri- angle whose vertices are p^ p^ p s . By Art. 17 the product obeys the associative law, so that Hence, by Art. 14, the product is what is generated by the point-vector p t p 2 in moving, || to itself, in the plane of the three points, from its original position, till it passes through p s ; that is, & parallelogram whose area is twice that of the con- necting triangle. Let p. 2 p\ = c, and p 3 p l = c' ; then PiP*Ps=Pil>*'=Pia' (39) The product is thus a posited and directed plane area of given magnitude; that is, a point-plane-vector, or simply a plane for brevity, especially when the magnitude is a matter of indiffer- ence. Eq. (39) shows that multiplying a plane-vector by a point fixes its position by making it pass through the point, since by Art. 14, ee' is a plane- vector. Let e" be a vector in any direction ; then, as in Art. 18, p^ + are" may be any point in space. Xow (?! + *")' =;?,' +o*"c f ; (40) so that the point-plane-vector obtained by multiplying ' by the point p l + a*" is, in general, different from p^t'. If, how- ever, we have e" = ye. + 2e', so that e" is || to ct', and p l + a*" is a point of the plane p\p. 2 p^ then ee' = ( Pl + x(y e + '))'=l>i'- (41) 28 DIRECTIONAL CALCULUS. [ART. 21. Hence two point-plane-vectors are equal when they have the same area and direction, or sense, and lie in the same plane, without regard to position in that plane. 21. Product of three vectors. Let vectors not || to the same plane ; then Art. 14, the product is the parallele- piped generated by the plane-vector e^a as it moves, || to itself, from the initial to the terminal point of e 3 . p Let e = c 2 , c 3 be any three = (.& c 3 , and, by then e, e', e" may be any three vectors whatever with suitable values of the scalar coefficients, and 1 2 3 2] 2 2 #3 (42) Hence two triple products of vectors can only differ in magni- tude and sign, and two such products will be equal when their magnitudes and order of generation, or sense, are the same. It follows therefore that the combinatory product of three vectors is always a scalar quantity, by the definition given in Art. 1. If the three vectors are parallel to one plane, the volume of the parallelepiped becomes zero, and the product therefore vanishes. Hence the planimetric combiwxtory product of three vectors is always zero. If we regard e 1? c 2 , c 3 as points at oo, we see that a point-plane-vector at GO is equivalent to a solid, which carries out the analogy mentioned at the end of Art. 19. * For the benefit of any reader who may not be familiar with deter- minants, it may be stated that the coefficient of eje.^ in (42) is an ab- breviated way of writing which expression will be obtained by multiplying out the values of and remembering that terms containing repeated factors vanish. CHAP. II.] MULTIPLICATION. 29 22. Product of four points. Let p^ p& p^, p^ be four unit points ; then, by Arts. 17 and 14, we have and the product is the volume generated by the point-plane- vector 2hP*P3 when it is moved || to itself from its initial posi- tion till it passes through p 4 ; that is, & parallelepiped, of which Pip?, p\Pv and pp+ are three conterminous edges. This volume is six times the connecting tetrahedron of the four points, which accords with Art. 16. (See figure of last Art.) If Pa Pi = <> Pa Pi= '> and p 4 p l = e", then PiP*P^=PiP*P&" =PiP*'*" = PI'C". . . (43) The point p l + a* -f- yt' + zt" may be any point whatever, with suitable values of x, y, z, and we have ( Pl + a j e + y e ' + c")'e"= S p 1 ee'c\ ... (44) so that any point whatever may be substituted for p^ in (43) without changing the value. By the above, and by the pre- ceding article, it appears that a product of four points is equal to the product of any other four points having the same mag- nitude and intrinsic sign, or order of generation : thus such a product is a scalar; and, in fact, differs in no manner what- ever from a product of three vectors which has the same mag- nitude and sign. We may therefore write #,'" = ee'e" ........... (45) If the four points are in one plane, the product is zero, because the volume of the connecting tetrahedron is zero. Thus the planimetric progressive product of four points is always zero. 23. The stereometric, progressive product of four or more vectors is always zero. But we may have e l e^e^ meaning the vector e l times the scalar c^s 6 ^ etc. The stereometric, progressive product of five or more points is always zero. But we may have such products as p t p a 30 DIRECTIONAL CALCULUS. [ART. 24. meaning the ordinary algebraic product of the point-vector p\p into the scalar 24. From the preceding articles we have the following conditions : is the condition that two points shall coincide. i> l p.,p.. = Q ......... (47) i is the condition that three points shall be in one right line. y,y,/':;y'4 = . .... (48) is the condition that four points shall be in one plane. e,t, = .......... (49) is the condition that two vectors shall be parallel. 1 o 3 =0 .......... (50) is the condition that three vectors shall be parallel to one plane. In the further development of the subject it will be con- venient to treat separately two- and three-dimensional space, considering the former first. PLAXIMETRIC PRODUCTS. 25. In two-dimensional, or plane, space two plane- vectors, or two point-plane-vectors, cannot differ from each other except in magnitude and sign, since both are restricted to one plane. Hence they become scalar quantities. Furthermore, by Arts. 19 and 20, the product of two vectors is now identical in meaning with the product of three points. Thus, if Pipi=f and p 3 p l = c', we have, in plane space, *,' = ' .......... (51) Plane space is the locus of all points dependent on three fixed reference points. We shall call these e , e,, e 2 , and shall always take the area <& as the unit of measure of area, when CHAP. IT.] MULTIPLICATION. 31 dealing with a point system. That is, we write always in plane space, W,>=1 ............ (52) This is a great practical convenience, and in no way affects the generality of results. The sides of the reference triangle taken around in order are called the complements of the opposite vertices; thus ^e.2 = (complement of e ) = \e ^o = (complement of e t ) = !! e v = (complement of e 2 ) = \e 2 (53) the vertical line before the point being called the sign of the complement. In dealing with a vector system we shall usually refer to a unit normal system of vectors tj, i,, as stated in Art. 12. We shall then have in plane space 1*=! ............. (54) 26. Let Pi = ? u< then pip 2 = m. 2 TOo TOO [e , e,, e.,] = |e, (55) The third and fourth members of (55) are simply different ways of expressing the second. The fourth member is espe- cially noticeable for its symmetry. Eq. (55) shows that any point-vector is expressible in terms of the sides of the reference triangle. 'ii ( ] (> Again, p^p^Ste TOO TO! ni., n ??i n (56) as will be found on multiplying out and putting e^e, = 1, by (52). 32 DIRECTIONAL CALCULUS. [ART. 27. 27. EXERCISES. 1. Find by eq. (47) the condition that the three points p+^ p + e 2 > P -f f shall be collinear, and illustrate geometrically. .4/1,9. e^ + c^g + e^ = 0. 2. Show that if a line be expressed in terms of the sides of the reference triangle, and the sum of the coefficients be zero, the line passes through the mean point of the reference points. 28. Since the product of three points obeys the associative law, it can be regarded as the product of a point into a point- vector, or of a point-vector into a point. Thus, if L = p->p 3 , PlPsPsPl'PiPs=PlL=P3PsPi = Lpl} ' ' (57) so that this product is commutative. 29. Product of two lines, or point-vectors. The products hitherto considered have all been progressive ; we now come to one which is regressive. Since two lines in plane space must intersect, they come under Art. 16, (6). Let the lines be L^ and L 2 , let p be their com- mon point, and let p 1 and p., be so taken that Li=p pi, and L 2 =p pt. We may also write, A = J>o ( Pi -Po), Lo = /) (p a -po) ; now the product of the vectors (Pip ) (Pz Po) is the area of the parallelogram on these vectors, and is scalar ; the product of the pomi-vectors should certainly give this result, and, in addition, the point fixed by them, viz. their intersection. This is in accordance with Art. 16, (6). The product (Pi-Po)(P2-Po) is equivalent to p pip 2 ; hence we may write L l L 2 =p p 1 -p p 2 =p p l p. 2 -p (58) It is to be carefully noted that the third member of eq. (58) is not derived from the second by interchanges, according to the asso- ciative and alternative laws, but is an independent expression, CHAP. II.] MULTIPLICATION. 33 which gives the meaning of the product of two lines. It may be regarded as a model form for the treatment of regressive prod- ucts. Thus, if AB and AC are any two quantities whose product is regressive, and if A is their common figure, we shall always have AB-AC=ABC-A ......... (59) We have accordingly LLi = p p 2 pop! = p pspi -P = - PoPiPz -Po = - LiLz, (60) or the product of two lines is non-commutative. 30. Product of a point and two lines. Let Z/j and L 2 be as in the last article, and p be some point ; then pL t L 2 = p p pi - p p 2 = p poptfs po = popiPa PPo- (61) PoPiPs can be placed first because it is a scalar. This is a mixed product, that of the two lines being regressive, and that of p into their common point, p , being progressive. Also, pLiL 2 = -2)L,L i = L,L l -p ...... (62) The period is necessary in the last member of this equation to preserve the meaning ; that is, the product of the points .LjZ/a an( l P' Without the period the expression would mean the line L 2 multiplied into the scalar L t p. 31. Product of three lines. Let L 1} L. 2 , L s loe three lines, and PD Pi, Ps their common points, and take scalar factors n 1} n 2 , n s so that Li^tiipzps, L 2 = n 2 p s pi, J^s = n sPiP-2'i then = - njn 2 n 3 p 2 p 3 - prfspz p l = W 1 w 2 w 3 (|) 1 |> 2 jp 3 ) 2 . . (63) It thus appears that in plane space lines obey the same laws of multiplication as points. 32. Let two points be given each as the common point of two lines, viz. p^ = Z/ ^i an( i P = L L 2) then L L 1 L. i 'L , ........ (64) 34 DIRECTIONAL CALCULUS. [ART. 33. which is a reciprocal equation to (58). Similarly, if we have P-2 = 2 . (65) 33. The condition ............ (66) requires that LI and L 2 shall be congruent, that is, be situated on the same right line; for, by (58), this gives The condition . .' ......... (67) requires that the three lines shall jxtss through a common point; for L 2 L 3 is some point, say p, and L^p = makes L^ pass through p, by eq. (47). 34. Product of parallel lines. Let the lines be L = p^ and L-2 = np.f. ; then L^ z = np^ - p.f mp! ep. 2 = ntpipz c = np^-f. e. . . (68) Thus the product is the common vector, or point at GO, mul- tiplied by the scalar 35. Product of two vectors. Let ti = m^i + then, since i^o = 1, (69, But, since e^ is a parallelogram whose sides are || to ^ and we have m ,.,= Te 1 7 Also, Wj = Tej cos < EI = Tt! cos a^ say ; m 2 = Tf l sin aj ; ri! = TV, cos a 2 ; H 2 = Ze 2 sin u 2 . CHAV. II. ] MULTIPLICATION. 35 Therefore which affords a proof of the trigonometrical formula for the sine of the difference of two angles. 36. Sum of fjoint-vectors. Using L L , L^p^p^pz as in Art. 29, we have LI + L, = PM + p Q p. z =p (P! + p 2 ) = 2p p, . . ( 71 ) if p is the mean point of 2h an( i Ps- Thus, L l -+- L 2 passes through the common point of the two lines and is equal in length to that diagonal of the parallelogram on L l and L 2 which passes through p (> . Similarly, so that the difference of L v and L 2 passes through p n and is equal in length to the other diagonal of the parallelogram. Suppose L! and L to be parallel, and equal respectively to npif and j>-f 5 then A + A = (^i + l> 2 ) = ('t + l)7'e; .... (73) that is, the sum is a || point-vector passing through the mean point of npi and p 2 . Finally, let n = 1, so that and the sum is, in this case, the product of two vectors ; that is, in plane space, a scalar. The reader will notice the exact correspondence between the results of this article and the resultant of forces in plane space. 37. Sum of sides of a polygon. Let 1, 2, 3 be any three points ; then 12 + 23 + 31 = 23 - 21 + 11 - 13 = 2(3 - 1) - 1(3-1) 36 DIKECTIONAL CALCULUS. [ART. 38. Hence the sum of the three point-vector sides of a triangle taken around in order is equal to twice the area of the triangle. Similarly, let 1, 2, 3, 4, n, be n points, the vertices of a poly- gon of n sides ; then 12 + 23 + 31 = (2 - 1) (3 - 1) 13 + 34 + 41 = (3 - 1) (4 - 1) -l)n + nl= [(n - 1)- and, adding, 12 + 23 + 34 + - -f n 1 = twice the area of the polygon. 38. COMPLEMENT. (a) The complement of a reference unit is the product of the other reference units, so taken that thex>roduct of the unit into its complement shall be positive unity. This definition is perfectly general, and applies to either a point or vector system in space of any number of dimensions. We have already had examples in eq. (53,) as e = &, whence == ^^^ = * (6) The complement of a scalar quantity is the quantity itself. Thus, \n = n ; {i^ = i^ = 1. (c) The complement of the product of several factors is equal to the product of the complements of the factors. Thus, [ (rip) = |nj p = nip ; jtjig = Itjjig. (d) The complement of the sum of several quantities is the sum of the complements of the quantities. 39. Complement in a plane vector system. Taking a unit normal system t,, 4, we have, according to the previous article, Complement of i 1? written ]i : = i?, for ij|ii = 1^2 = 1 ; Complement of i% written J^ = i 1? for ijij = I 2 t 1 =t 1 i 2 =l ; II*i = 1*2 = .*i ; 11*3 = |*i = - *. CHAP. II.] MULTIPLICATION. 37 Let EI = 7H.ii! -f- and 2 = i h + n-ih. j then By the figure it is evident that J X is 5*j a vector of the same length as e^ and perpendicular to it, or, in other words, taking the complement of a vector in plane space rotates it posi- tively through 90. The product e^ is the parallelogram whose sides are cj and [c 2 ; if tj is parallel to je^ the area of the parallelogram van- ishes, or cjje, = ; but, since |c 2 is _L to c a q must, in this case, be _L to c 2 ; hence the equation i!*-0 (76) is the condition that the two vectors ^ and e, shall be perpendic- ular. The product e^ is the area of a square each side of which is of the length T^ ; hence -I- 7 12 * e 2 CO-IT (T7\ e ii e i * i r5 sa y i. * * ) The form e^ is merely another way of writing e^ which is often convenient. If n is a scalar, we have nn = nn = whence the analogy is apparent. Grassmann calls e^ the " inner product " of ei and c 2 , regard- ing the complement sign as a species of multiplication sign, and accordingly calls ^ the " inner square " of e. It seems prefer- able to the author not to introduce a new species of multipli- cation, but to regard 1 'e 2 as simply the combinatory product of cj into Icj, a way of looking at it which is practically far more simple, and renders interpretation easier. Somewhat after the analogy of the word cosine for sine of the complement, we may call ^ the co-square of e, and, just as we read a 2 , a square, we may read c?, e co-square. Similarly, c a ]e 2 may be called the co-product of ej and 3. 48607 38 DIRECTIONAL CALCULUS. [Aux. 39. We have also, e,fc 2 = (m^! -J- ?n 2 i 2 )| ( n i l i + 7| i; l :>) = wVi 4- wi 2 M 2 = f-2 'i- (^8) We obtain the third member because *ll l l = 1 1 1 2 = 1> I l! l 2 = l l'l = 0, I 2 'li = 1 2 1 2 = 0, and iz\i 2 = toti = iii 2 = 1 ; the fourth member is apparent from the symmetry of the third. It will be found that we have always, when A and Z> are quan- tities of the same order in the reference units, A\B = B\A ........... (79) Since Tie = T e, as shown above, we have, as in Art. 35, ejcs = T I Te 2 x sin (ang. bet. e, and ],) = 7^ Te., cos < c l (80) Also, taking the values of m 1} m 2 , i], w 2 as i n -A- 1 ^- 35, we have tjjc, = m^i + wi 2 n a = Tcj Te 2 (cos aj cos a., + sin a! sin a.,) = r I 3 r7 2 cos< e -, ....... (81) which affords a proof of the trigonometrical formula for the cosine of the difference of two angles. If c 2 = EU we have * ' Square and add eqs. (70) and (80) ; therefore r 2 i r 2 2 = 1 i 2 i = ( 1 2 ) 2 +( 1 | 2 ) 2 . . . . (83) If i be any unit vector, and p any other vector, we have t p i = i Tp- cos<^ = projection of p on direction of t, and ip \L =\t Tp sin<^ = projection of p on direction X "to t; hence we may write p = i p|i + ip - i ......... (84) This equation may be verified by multiplying successively by t and |i. CHAP. II.] MULTIPLICATION. 39 40. Complement in a plane point system. Taking as reference points e , e l5 e. 2 , with the condition e^e^ = 1, we have, in accord- ance with Art. 38, = (85) = K-K. e,, by Art. 38, (c), - O]^O9 * OQOQ ' ^Q^l ( ^0^1^2/ *""~ ' Note that the complement of the complement of a reference point is here the point itself, while in the vector system the complement of the complement was negative. The general law is that when the number of reference units is even, the com- plement of the complement of one of them is negative, and when the number is odd, it is positive. two points ; then _ _ n (, i \ n g o i >e \ . (86) so that the complement of any point is a point-vector. The fourth member of (86) expresses this point-vector as the prod- uct of the points in which it cuts two of the sides of the refer- ence triangle, so that it may be easily constructed. We have also so that eq. (79) is here verified. Note that the product is scalar. We have, however, .,-... (88) and ei|eo e i = e^, = |e ) so that, when the quantities on each side of the complement sign are of different order, the product is not scalar, nor commu- tative about the sign. 40 DIRECTIONAL CALCULUS. [ART. 41. If LI=\PI and L 2 = \p 2 , we have pi=\L l and ^ 2 = \L 2) and L,\L, = |l>i \ Ps = \ Pl p 2 =p.^p l = Pl \ Pa = L 2 \L,, (89) which also agrees with eq. (79). Again, L,L,=\ih'\P>> = \lhIh\ (90) so that the common point of the lines L and L., is the comple- ment of the line PlPi . Also, L ] \p 2 = ^(L ] \p 2 ) = \(\L l -p 2 ) = \(-p 2 \L l ) = -\(p 2 \L 1 ), . (91) of which eqs. (88) are a special case. 41. Geometric interpretation. For the sake of simplicity, suppose e , e 1} e 2 to be the cor- ners of an equilateral triangle. With e, the mean point of P the triangle, as a center, draw a circle of radius = , a being the VG side of the triangle; then, p being any point whatever, \p is its anti-polar with reference to this circle ; that is, a line par- allel to its polar, and equidistant Avith it from the center of the circle, but on the opposite side. It is evident that with reference to the above circle each vertex is the anti-pole of the opposite side of the reference triangle ; for the respective distances of a vertex and its oppo- site side from the center are aV3 and - V3 3 2 2V3 and the radius of the circle is a mean proportional between these. The figure shows the construction for the radius of the circle, and also for the anti-polar of any point p. If the reference triangle is not equilateral, it can be obtained by projection from an equilateral one, and the circle corre- CHAP. II.] MULTIPLICATION. 41 spending to that will be projected into an ellipse such that, with reference to it, \p is always the anti-polar of p. Hence it will only be necessary to prove the property in the case of an equilateral reference triangle. Before proceeding to the proof, however, it will be necessary to find an expression for the dis- tance between two points in terms of their coefficients. 42. Distance between two points. Let the points be p^ and p 2 as in Art. 40. Since they are unit points, we have 2^ = 1 = 2iWi ; .'. In = 1 /i Z 2 , m n = 1 m l m 2 . Hence and p. 2 = 2we = e + tn>\(e\ e o) + m. 2 (e. 2 e ) = e {] -f- m^ -}- m&. Thus, p. 2 pi = (wj ?0 fi + ( w 2 Z 2 ) c 2 , and the required distance by eq. (82), (92) In the case of an equilateral reference triangle, whose side is a, we have e/ = e 2 - = <* 2 > and ! c 2 = a 2 cos 60 = a 2 ; so that eq. (92) becomes 2 -Z 2 ). (93) 43. Proof of the anti-polar property of the complement in a point system. Referring to the figure of Art. 41, we propose to show that which will establish the proposition. We have, since p' is the common point of \p and pe, p' congruent with pe ' \p = xp + ye. 42 DIRECTIONAL CALCULUS. [ART. 43. Multiply both members by pp^ ; then, ppi ' p& ' \P = ppfi ' P\P = ypp^ '> whence y = p-- Multiply the same equation by p,e ; therefore p^e -pe-\p ep,p e\p = whence x = e\p. Substituting, we have p- p e Hence we have r e e\p p e\p-(p * ep' = e J ff - = |J ^ _ p- e\p p- e\p and pt e\p Now, by eq. (93), taking Z , l>i, k as the coefficients forp, and m = m l = m 2 = ^ as those for e, we have Also, p*=p\p = tf and e|p = K^o + i i + ^) = i' so that & - e\p = 2(1? + I? -I,- I, + 1,1, + 1). Hence (95) becomes - e) T(e -p>) = * a ^ + *** + ?1/2 ~ Z ' " ?2 + i> =^- Q.E.D. CHAP. II.] MULTIPLICATION. 43 44. The conception of the complement in a point system, as developed Arts. 40-43, is not found in Grassmann's works. He deals exclusively with the complement in a unit normal vector system. See Die Ausdehnungslehre, 1862, Art. 330. In a remark at the end of Art. 337 he shows how the idea of the complement might be extended to a point system, but in a way entirely different from mine, and one which he himself evidently considered of no practical value, since he has made no application of it. On the contrary, the method above de- veloped is of great utility, giving at once reciprocal properties, according to the principle of duality. In a subsequent chapter the anti-polar property of the com- plement will be established in a different manner, directly, for a reference triangle of any shape. 45. A multiplication table for a point system in plane space is given on page 44. The product of any quantity at the left into any at the top is found at the intersection of the corre- sponding row and column. Thus, e.^e efa = e& e 2 \e = 0, etc. Algebraically considered this system forms a seven-fold alge- bra, seven reference quantities being required to express all quantities of the system, including scalars. 46. Projections. If we write the equation P = Xfy + x.fy ........ . (96) Xfa is evidently the projection of p on 1 \\ to e 2 , and x.f. 2 is the projection of p on c 2 || to c^ Multiply / (96) into c 2 ; therefore pc 2 = x^e^ since / pe, e-jCa = ; thus, x l = -. Similarly, mul- 1 C 2 P e i tiplying into e l5 we have x 2 = - ; whence '-o e , c i p = li-^- 2 + ^l 1 . ....... (97) Hence the projections of p on e l || to c 2 and on e. 2 \\ to cj are 44 DIRECTIONAL CALCULUS. [ART. 46. A MULTIPLICATION TABLE FOR A POINT SYSTEM IN PLANE SPACE. II II CHAP. II.] MULTIPLICATION. 45 respectively - - and - - -. These are particular cases of iz si a general proposition which may be stated as follows. Let B be a quantity of the nth order in the reference units, and C one of the mth order, and let the number of reference units be m + n, so that BC is scalar : then the projection on B of any quantity A, directed by C, is B-AC BC ' Similarly, the projection of A on "(7, directed by B, is C-AB CB We shall have also A _B.AC . C.AB ~~ CB Since we have restricted ourselves to space of two and three dimensions, we shall not give a general proof, but shall verify and explain the proposition in such cases as arise under this restriction. See Die Ausdehnungslehre, 1844, Chap. 5, and the same, 1862, 127-129. If in (97) EU c 2 are a unit normal system, replace them by ^ and i 2 ; then, since i^ = 1, p = ti pt-2 2 P l \ = i P! I I + IIP 4 u .... (99) which agrees with eq. (84). 47. Consider next the point equation P = XnPo + x\p\ + x?Pz 5 ........ (100) multiply successively by pip 2J p 2 po, and p tt pi, and we find pPiPs PPzPo PPoPi whence p = - - [ p - pp&t + Pi pp 2 po + Ps PPoPi] - 46 DIRECTIONAL CALCULUS. [ART. 47. Each, term in the brackets taken with the outside factor is of the typical form (98), and is the projection of p on one of the points on which it depends. Write again, P = and multiply into \p^ \p 1} \p 2 successively, and we find PoPiP? ' ( 102 ) We might also have obtained (102) from (101) by putting \pip 2 for PQ, etc. The terms of the right-hand member of (102) are again of the typical form, and are the projections of p on the anti-poles of Pip 2 , P*PO, and p pi. Note that \PlPa ' \Po=\(PlPaP*) = PlPaPo = PoPlP2- If in (101) we take the last two points together, their sum is some point on p t p 2 and also on pp^ since p is expressed in terms of p and this point. Hence this point is congruent with PPo> an( l we ma y write or Miiltiply into p\p& and we have PPiPa = vpoPiP* Multiply into p pi, and we have PPoPi yPiPz 'PPo 'PoPi = VPiPaPo -PPoPi, Substituting values of x and y, we obtain PiPa'PPo _ , an equation of the same form as (98). The second term of the right-hand member of (103) is the projection of p directed by p& CHAP. II.] MULTIPLICATION. 47 The second members of (101) and (103) must be identically equal ; hence we have PiPa PPo = Pi PPaPo + Ps ' PPoPi, or, writing p s instead of p, and p^ instead of p , for symmetry, PiP2 PsP* = -Pi- P 2 PsP4 + P 2 PsP*Pi I 1()4 , = Ps PtPiPz P* PiPzPa f ' The last expression is obtained by interchanging the suf- fixes 1 and 2 with 3 and 4. If in (101) and (102) we put the reference points e , e ly e 2 for the p's, the equations become iden- tical, viz., j> = o- J pe<> + i-.pK4-j-p|s ....... (105) Similarly, (103) becomes p = e 'p\e +\e -pe n = e 'p\e -{-e p-\e , . . . (106) a form analogous to eq. (99), expressing p in terms of its pro- jections on e and e^. 48. The operations of the last article would have been pre- cisely the same if lines (point-vectors) had been used through- out instead of points. Hence, substituting L's for p's in eqs. (101) to (106), we have L = 1 [7,o LL,L, + L, LL 2 L + L, LL L,l . (107) JjQ^L/2 L = - [|AA L L +\L 2 L n -LL,-}- \L L 1 - L\L.^ (108) \LZ jjp.Li ^ LI ' L^L^L^ -f- L. 2 L = e^., - Le Q + e. 2 e Le + e e i Le 2 , ...... (HI) L = e^ Le u -j- e Le v e 2 = e^ L\e^ -{-(e^ L) e-^e.,. (112) 48 DIRECTIONAL CALCULUS. [ART. 49. In equations (107), (108), and (111) L appears as equal to the sum of its projections on three given lines ; in (109) and (112) it appears as the sum of its projections on a line and point. The projection on the point in (109), viz. 1 2 ' - ? , LiL 2 L is a certain portion of the line joining the common point of L t and L 2 with the common point of L and L , the reciprocal idea to that of the projection of p on pip 2 in (103). 49. In eq. (104) put p 3 p 4 = \q 1} and we have PiPt'\Qi = -Pi-Pa9i+Pa'PiQi ...... (H3) In (110) put L 3 L 4 = MI, and we have L t L 2 M l = - L, L 2 M l + L, -L.M^ . . . (114) Note that q l is a point, the anti-pole of p s p 4 , and M l is a line, the anti-polar of the point L 3 L 4 . Again, in (110) put L 1 L 2 = p 2 , L 3 = \QI, L+ = \q 2 ; then, since ! . 2 = , we have and similarly from (104), l ..... (116) Multiply (115) and (116) respectively by p^ and LI, or (113) and (114) respectively into* |0 2 and \M 2 , and we have (118) If 0i =PI and Q 2 = p 2 , Mi = A and M 2 = L% (117) and (118) become P\Pz\P\Pz = (P\P2)~ = Pi~P2- (Pi\P2) 2 ) /im\ 2 I ' ' (11") * ^1 multiplied &/ .B means BA ; A multiplied into B means AB. CHAP. II.] MULTIPLICATION. 49 Put q 2 for p in (102), and multiply by p pip 2 -i Let E! and 2 be || and equal to two of the sides ; then 2 e t is || and equal to the other side. Let p 1 e (> = x^ and lh e o = ^2*2- Then, by given condi- tions, x 2 e 2 a^ = n (e 2 E!) . Multiply by e x and 2 successively, and we have * A treatment of the matter will be found in the fourth chapter of the Ausdehnungslehre of 1844, and a more extended one in an article by the Author in No. 1, Vol. IV. of the "Annals of Mathematics," published at the University of Virginia. CHAP. II.] MULTIPLICATION. 51 (3) The bisectrix of an angle of a triangle divides the oppo- site side into segments proportional to the adjacent sides. With the figure of the last proposition the bisectrix of the angle at e is e^e^Te* e. 2 T 1 ), which gives the proof imme- diately, for the point e^ Te 2 e 2 T^ is on the line efa at dis- tances from these two points inversely as the weights, i.e. directly as 2V and Te 2 , and between them or outside according as we use the upper or lower sign. (4) If a, b, c are the three sides of a triangle, to show that st a 2 = b' 2 -\- c 2 2 be cos < , With the figure above let T( 2 - l )=a, T! = C, Tt 2 =b. Then, a 2 = T 2 (e 2 - l ) = (e 2 - l ) 2 = ^ + r - 2 l |e 2 = b 2 -f c 2 2 be cos < , (5) Find the condition that lines through the three vertices of a triangle shall have a common point. By the figure the condition is = 0. = n^ + Z w e 2 ) Let then, This is equivalent to Pe ' Pie P-i - eiPo ' e*Pi e p 2 = 0. (6) Find the condition that three points on the respective sides of a triangle shall be collinear. This case is the recip- 52 DIRECTIONAL CALCULUS. [ART. 51. rocal of the preceding. Let the points be as in Ex. (5) . Then the condition is = = (mfa + n e 2 ) (^2 + ^o) (l&o + = ra n 1 Z2+noZ 1 m 2 , which is equivalent to Po2 P& Ptfi + Wo ^Pi eoPi = 0. (7) Show, by Ex. (5), that the following sets of lines in a triangle have a common point. 1st. Lines through the vertices and the middle points of the opposite sides. 2d. A line through one vertex and the middle of the oppo- site side, and two lines through the other vertices || to the sides opposite to them. 3d. The bisectors of the angles ; all internal, or one internal and two external. 4th. The perpendiculars from the vertices on the opposite sides. 5th. The perpendiculars to the sides at their middle points. (8) Show, by Ex. (6), that the points where the bisectors of the angles of a triangle cut the opposite sides are collinear, if two of them are internal and one external, or if all are external. (9) From the values of x^ o^, x^ given just before eq. (101), determine the effect upon the position of p of giving a negative value to one or more of these coefficients. (10) If in the result of Ex. (6) lines be substituted for points, say (L^, LI, L 2 ) for (e^, e^ e 2 ) and (L ', LI, L 2 ) for (Poj Pi>Pz)> interpret the resulting equation. (11) If a quadrilateral be divided by a right line into two quadrilaterals, show that the common points of the three pairs of diagonals are collinear. CHAP. II.] MULTIPLICATION. 53 (12) By substituting lines for points, in the equation of condition of the last exercise, derive the reciprocal proposition. (13) If two triangles are so situated that the lines joining their vertices two by two meet in a point, then will their cor- responding sides meet each other in three points lying in one right line. (14) Show that, if a line L cut the six lines that can be drawn through four points e 1} e 2 , e s , e^ in the six points Pi, P Ps, Pi, Ps, Pat as in the figure ; then the relation Tp 3Pl ' . Tp 2 'p 3 ' = T Pl 'p a ' Tp 3 ' Pl . T PaPa holds. These points are said to be in involution. (15) By substituting lines for points, and a point p for L, obtain the recipro- cal theorem, and interpret it. (16) Let PD p 2 , p 3 , p^ be four fixed points, and let p 2 ' and p 3 ' vary subject to the conditions = 0; find the locus of p, the common point of p 2 p 3 ' and p 2 'p 3 . If pip 2 p s = Q, show, by eq. (104), that the locus becomes two straight lines, one of which passes through p 4 . We have at once p 2 = pp 3 p 2 p 4 and p s ' = pp 2 p 3 p+ ; whence, by substitution in above condition, we have Pl(PP*'P*P4) (PP2 = 0, the equation of the locus, which, being of the second degree in p, represents a conic. On applying (104), this will separate into two factors of the first degree in p, if p^p 2 p s = 0. 54 DIRECTIONAL CALCULUS. [ART. 51. (17) Interpret the reciprocal results obtained by putting L's for p's. (18) If e lf e. 2) e 3 , e 4 are four coplanar points, and e 5 and e 6 are the common points of e^ and e s e 4 , and of e t e l and e 2 e s respec- tively, show that the middle points of e^, e 2 e 4 , and e 5 e G are collinear. (19) Lines through the vertices of any triangle and the corresponding vertices of its complementary triangle meet in a point ; and, reciprocally, the corresponding sides cut each other in three collinear points. (20) A triangle whose sides are of constant length moves so that two of its vertices remain on two fixed straight lines : find the locus of the other vertex. Let e c i an d e c 2 be the two fixed lines, and Pip-^p the triangle. Also let p l e = jctj and p 2 e = ye 2 ; then p 2 p l =. ye 2 ajcj, and we have the con- 6 o dition - 3* = c. Let pe be _L to p 1 p 2 , Tp^ = me, Tep = nc ; then, n\ (ye 2 xej. This equation in p and the scalar variables x and y, with the condition above, which is really of the second degree in x and y, is that of a conic section, which must evidently be an ellipse. The student should eliminate x and y by multiplying succes- sively by ej and e 2 , thus obtaining a scalar equation in p of the second degree. (21) Show that the expression ppiL-ip 2 L 2 p^p [ ', interpreted according to Art. 16, (c), is identically equal to and from this that it is also equal to CHAP. II.] By eq. (104), MULTIPLICATION. 55 3 'p'L a p' = (PPi Li'PaPa) -P'L-2 - (PPi ' A 'PsP') = (PiPaPa -pLi PPsP* 'PiLi) 'P'L* - (PiPsP 1 -pLi pp 2 p' -piA) ' Pal's = (Pi -PL, -p -p l L l )p 2 (p 3 -p'L 2 -p' The second part is left to the student. (22) If, as in eq. (86), 1 ^0 show that /TT| __ fJlT __ -~ / / 7 7 \ 2 2 | / 7 7 \ 2 2 O / 1 1 \ i 1 7 \ I J. I /_/ .i J-^ ^v ( t-Q "" tj I 2" ~T" \ 'o ^~" ^2 / ^1~ ^^ V "^ I/ V "" "~ ) ^1 ^* in which q = e 1 e and e. 2 = e 2 e . (23) If L 1 =p 1 p 1 ', L 2 =p 2 p 2 ', L 3 =p 3 p s ', then show that P 2 L 3 P 2 'L S ~ pJL, p, 1 . (24) By eq. (119) prove that u'Li p 3 'L 2 2/2 (25) Show by eq. (120) how the product of two determi- nants of the third order may be expressed as a determinant of the same order. (26) If A = 2o(Z|e) and L a =$(m\e), show that when m = 0, then LI and L 2 are parallel. 56 DIRECTIONAL CALCULUS. [ART. 52. STEREOMETRIC PRODUCTS. 52. Three-dimensional space is the locus of all points dependent on four fixed points. Let these four reference points be e , e 1} e 2 , e s , so situated relatively to each other that e^e-fts = 1, always ; i.e. the unit of volume is six times the volume of the reference tetraedron. Let four points be taken, viz. : p 1 = Sofce, p 2 = Sjfe, p 3 = Some, p^ 5^M ; then, = A = KQ KI 1 I, 1 2 3 to li 1 2 l s K 2 K s A-j [e , 61, 62, c 3 ]. (123) The first result will be obtained by actual multiplication of the values of p t and p 2 , and the second result is simply an abbreviated way of writing the other as in eq. (55). It ap- pears thus that any point-vector in space is expressible in terms of the six edges of the reference tetraedron. Again, 1 2 3 LQ LI /2 '3 1 2 3 ^o n '2 ^s ra m, wi., m 3 (124) in which the third member means the sum of the four third- order determinants that can be formed of the columns taken three at a time, each multiplied into its corresponding triple product of the reference points, with the same order of suf- fixes. In the fourth member |e =e 1 e 2 e 3 , |ei= e 2 e 3 e , |e 2 = e 3 e o e i> and \e s = e^^. Thus any point-plane-vector is expressible CHAP. II.] MULTIPLICATION. 57 in terms of the four faces of the reference tetraedron. Of course the product of three points is not scalar in solid space. Finally, / (125) m because e^e^ = 1 ; thus the product of four points is scalar, as was shown in Art. 22. As an exercise let the student find the condition that the plane P^ in (124), shall pass through the mean of the refer- ence points. 53. Since, by Art. 17, the continued product of four points obeys the associative law, we have = PiPn = -p 2 . Then, PjP 2 = Lp 1 Lp. 2 = Lp,p 2 -L > P t P 1 = l4> s -l4> 1 = Lp a p 1 .L = -P l P a ) 5 so that the product of two planes, like that of two points, is non-commutative. If P! and P 2 are parallel, L is at oc and becomes a plane- vector ; call it t]j and substitute in (130) ; then we have for the product of two planes, having a common line t] at oc, . . (131) 56. Product of three planes. Let p be the common point of PI, Pa, and P 3 , and take p lf p 2 , p s on the common lines of these planes, so that then, PiPsPa = Pop a p 9 poPsPi PoPiPa = 023 013 . 012 = -023.0132.01 = 0123.023.01 (132) = 0123 . 0231 = (jpoPiM.) 2 -l>o In this equation we have used for p , 1 for p^ etc., for con- venience. This we may frequently do when no ambiguity will result. In eq. (132) we have worked according to Art. 16, (c), by which P 1 P 2 P S = P l P 2 P 3 ; but if we had combined P! and P 2 first, and the result with P 3 , we should have obtained the same result. Hence planes obey the same laws * of multi- plication as points, in solid space. * We have here assumed the distributive law to hold, as, in fact, it does, for all products, progressive, regressive, or mixed ; but it is easy to prove the law for planes or lines, assuming it to be true for points. Thus, taking the planes as above, P,P, + PjP 3 = 023 031 + 023 . 012 = 0123 (03 - 02), because 0123 is scalar, = 0123 0(3 - 2) = 01 (2 - 3)3 0(3 - 2) = 0(3 - 2)13 0(3 - 2) = 0(3 - 2) 1 0(3 - 2)3 = 023 01(2 - 3) = 023 . (031 + 012) = P,(P, + P 3 ). CHAP. II.] MULTIPLICATION. 59 If the three planes are parallel to one right line, the common point is at GO, and e may be substituted for p Q in (132). 57. Product of four planes. Let p^ p. 2 , p s , p be the four common points of four planes P^ P a P 3 , P 4 taken three by three, and take four coefficients n 1} n^ so that P 1 = ihp a p 3 p^ etc. ; then P!P 2 P 3 P 4 = W i 3 tt 4 . 234 . 341 . 412 - 123 | = n l n 2 n 3 n 4 (2hp-2p 3 p4Y > > Mixed products are to be interpreted according to Art. 16, (c). Thus, has this meaning. P?L S is a point; this, multiplied by L& gives a plane ; this, by P l} a line ; and this, by LI, a scalar quantity. 58. Products of plane-vectors. Let 7/1 and 7/ 2 be two plane- vectors (lines at oo ), and let c be parallel to each of them, while Ci and e 2 are so taken that 7/1 = eti and i) 2 = ee 2 ; then, This result may be obtained directly from eq. (58) by regarding the points and lines of that equation as all at oo, and therefore necessarily in the plane at oc. The product of two plane-vectors appears as a vector parallel to each of them, multiplied by a scalar quantity. We have at once Next take a third plane-vector j/ 3 , and let e t be || to rj 2 and rj s , e 2 || to 773 and 7/1, e 3 || to 7/1 and 7/ 2 , while the tensors of l} etc., are such that 7/1 = e>e 3 , rj.> = e 3 j, 7/ 3 = e^o > then As an exercise let the student discuss yP, the product of a plane-vector and a point-plane-vector. 60 DIRECTIONAL CALCULUS. [ART. 50. 59. Equations of condition. By eq. (48), pP=0 (137) makes p lie on P, or P pass through p ; L 1 L 2 = Q (138) makes the two lines intersect ; LP=0 (139) makes L lie in P, or P pass through L ; AP^O (140) makes the two planes coincide ; P!P 2 P 3 = (141) makes the three planes pass through a common line ; for P 2 P 3 is a line, say L, and, by (139), P,Z,=0 makes P! pass through L; PaP 2 P 3 P 4 = (142) makes the four planes pass through one point, for PjPgPg is some point, say p, and pP 4 , by (137), makes P 4 pass through p ; PjLP^O (143) makes Pj and P 2 cut L at the same point ; for, writing L = PgP,, the result follows from (142) ; W = (144) makes the two plane-vectors parallel ; W72>?3 = (145) makes the three plane-vectors all parallel to one straight line. Equations (138), (140), (141), (142), (144), (145), should be compared with equations (66), (46), (50), respectively. 60. Addition of planes and plane-vectors. Let P^ and P 2 be two planes intersecting in L, and let p t and p 2 be so taken that P! = Lp l and P 2 = Lp 2 ; then P! +P 2 = L Pl + Lp, = L( Pl +p,) = 2Lp, (146) in which p is the mean of p l and p 2 . Thus the sum is that CHAP. II.] MULTIPLICATION. 61 diagonal plane of the parallelepiped, of which two adjacent faces are Pj and P 2 , which passes through L; the parallelo- grams P! and P 2 being so placed as to have a common side L. If the t' vo planes are parallel, let rj be a plane-vector parallel to each of them, i.e. their common line at oo, and let p l and p 2 be points of the respective planes ; then we may write P 1 = n^r] and P., = n 2 p 2 rj ; whence Pj + P 2 = n l r 1 p l + n 2 r)p 2 = rj (n^ + n 2 p 2 ) = i 277 If HI + n 2 = 0, then (148) so that the sum, in this case, becomes a volume, and is scalar. Cf. eq. (74). Take P u P 2 , P 3 , p > Pi> Pv Ps as in Art. 56 ; then Thus the sum is a plane through the common point parallel to the plane p\p 2 p z - If p is at oo, call it c ; then each plane is || to c, and the sum becomes the product of three vectors, and therefore scalar. If P 1 + P 2 = 0, or P 1 = -P 2 , ..... (150) the two planes are coincident. If P 1 + P 2 + P 3 = 0, ........ (151) the three planes pass through one right line, as appears by comparison with eq. (146) . Similarly, P l + P 2 + P 3 +P 4 = ....... (152) causes the four planes to pass through a common point, as appears from eq. (149). Take rji, r/ 2 , e, cj, e 2 as in Art. 58 ; then >7i + >?2 = i + 2 = e( e i + c 2 ), .... (153) so that the sum is a plane-vector parallel to c. 62 DIRECTIONAL CALCULUS. [Aur. 01. 61. Addition of point-vectors, or lines. Take n point-vectors Pi f D P2 f 2> ' 'Pn f n) an( l ca ll their sum S ; then ' It appears that /S is, in general, composed of two parts, of which one is a point-vector, and the other a pZome-vector. If this plane-vector is parallel to the point-vector, i.e. capable of expression as the product of some vector a into 2, then their sum can be expressed as a point-vector only ; for we have, in this case, S=e 2e + a2c = 0o-f a)2e, a point-vector of the same length as e^e, \\ to it, and distant ^ from it by the amount To. sin < e - S being composed of two parts which cannot be equal to each other, if we have the equation S = 0, it can only be satis- fied by making each part separately zero, so that S = implies 2e = and 2(p e )e = 0. The quantity S maybe called a screw,* and we shall hereafter consider some of its properties. 62. TJie complement in three-dimensional space. Following the definitions of Art. 38, we have for a unit normal vector system ij, t 2 , i 3 , Let cj = Z^ + l. 2 L 2 + Z and c 2 = m i l i + then |i=^i 2 t 3+Wi+^ t i'2 = 7 (li<*t k l i)(li<-3 ^1)5 * See "The Theory of Screws," by R. S. Ball, Dublin, Hodges, Foster & Co. ; and also a paper by the author on "The Directional Theory of Screws," Annals of Mathematics, Vol. IV., No. 5. CHAP. II.] MULTIPLIC ATION. 63 so that e l is a. plane-vector. The third member of (156) is the product of two vectors ; the first, ^t 2 l-*i, is easily seen, by the figure, to be _L to l^ -f- ^ 2 > the projection of q on the plane -i iji,, and hence _L also to ej, because _L to the plane that projects Ci on iji 2 ; similarly, l^ Z 3 i x is _L to Z x i x + Z 3 t 3 , hence to the plane that projects e x on i 3 i x , and therefore to e x itself. Hence |e x is a plane- vector perpendicular to e x . Since ||e x = q, it follows that the converse is true ; that is, the complement of a plane-vector is a line-vector perpendicular to it. It is evident from the figure that c x is a diagonal of the rectangular parallelepiped whose edges are Zj, Z 2 , l s in length ; hence, T I = VV + tf + tf. (157) Multiply (156) by tj ; therefore e iei = l * = ^ + ^ + Z 3 3=T* ei , .... (158) so that, as in plane space, the co-square of a vector is equal to the square of its tensor. The product e^q is that of the vector E! into a _L plane-vector, as has just been shown ; it is therefore 64 DIRECTIONAL CALCULUS. [ART. 63. a volume which is equivalent to Ttj times the area of ej ; hence, by (158), the area of |c t is numerically equal to Te 1} or T\^ = Tfy' ........... (159) Thus the complement of a vector in solid space is a perpendic- ular plane-vector having the same tensor. We have j / . (Io9) I 3 m 3 = e 2 |e 1 ) Now e 1 |e 2) being the product of the vector e, into the plane-vector |e 2 , is equivalent to T|e 2 sin < Cz = Tej Te 2 cos < that is, (160) If cj and c 2 were unit vectors, 1 1} 1 2 , 1 3 , m^ m 2 , m 3 would be direction cosines, and thus (160) gives a proof of the formula for the cosine of the angle between two lines in terms of the direction cosines of the lines. By (160) the condition that e 1 and c 2 shall be at right angles Cl c 2 = (161) Let r)i = \*i and rj 2 = |c 2 ; then 7?! 77 2 ^ Ci C 2 ^^ C 2 Cj zzz Ci C 2 ^= J. ^JL 2 COS <^ ^^ J. 7?j J. 71 2 COS <^ . ( 1 () ) e l ^l and -qi\r]z = (163) is the condition of perpendicularity of two plane-vectors. 63. Complement in a point system in three-dimensional space. Let e , e 1} e% e 3 be four unit reference points, so taken that the product 606,6263 = 1 ; then I it j i 1 OQ 1 2>? I! 0""~ 123 "~~" ~~* 0? 1 ^~ ^2 3? ^01"^ 2 3^~ 0^1? |^1 == "~~ ^2^3^0? 11^1^ "~ ^2^3^0 == ^1? ^0^2 == ^3^1? 1 1 ^0^2 ^ | ^3^1 == ^0^2? |^2^ 6360^], ||^2 == I^S^O^l 213 ^2? |^0^3 == ^1^2J jl^O^S" &1&2^^&$3' 10 > O > \\O ^~ tf 3 o"l^2? 1 1 "3 CHAP. II.] MULTIPLICATION. 65 Note that the complement of the complement of a reference point is the point with negative sign, but that the complement of the complement of a reference line, or edge of the reference tetraedron, is the line with positive sign. We have \e fa -\e. 2 = e^Cg e 2 e 3 e e & e^ which agrees with Art. 38, (c). Let Pi = 2,lke, and p 2 = 2o?e ; then + etc., so that the complement of any point is a point-plane-vector, or plane, and any plane may be expressed in terms of the four faces of the reference tetraedron. From eq. (123) it follows that the complement of any point- vector or line is another point-vector. Again, Pi\Ps = 2fce| 2Ze = Z m -f- l^ -f I 2 m 2 + I 3 m s = p a \pi. . (165) Let P! = \PH P 2 = \p 2 5 therefore, Pl\P 2 =\Pl'\\P 2 =-\Pl'P 2 =P-2\Pl=Pl\P2=P2\Pl- (166) Let LI = fc^i + &26o6 2 + ^36063 -f- fc/e 2 e 3 + ^'636! -f A: 3 'e 1 e 3 and L 2 = l^e^ + l^fa + etc. ; then L 1 \L 2 =Jc 1 l 1 -\-k 2 l 2 +k 3 l s +Jc 1 'l 1 '+k 2 'l 2 '+k 3 'l 3 '=L 2 \L l . (167) Also L 1 L 2 = k l l 1 ' + k 2 l. i l + kJ s ' + k l 'l l + k 2 'l 2 + k 3 'l 3 . . (168) If L 2 = L l} we have L l 2 = 2(k l k 1 ' + k 2 k 2 ' + k 3 k 3 ') ........ (169) But if LI is a point-vector, its square must be zero, and as the second member of (169) is not necessarily zero, it follows that LI and L 2 are not, in general, point-vectors ; in fact, they are screws, as shown in Art. 61. We have then for the condi- tion that LI shall be a point-vector, kM + kJcJ + kJc^Q .......... (170) From (165), (166), and (167) it appears that a co-product in which the factors on opposite sides of the sign are of the 66 DIRECTIONAL CALCULUS. [ART. 64. same order is commutative about that sign, and always scalar. If the factors are not of the same order, this is not the case ; for example, P\L = -\\(P\L) = -\(\P-L) = -\(L\P). . . (171) Proceeding in a similar manner to that of Art. 43, it may be easily shown that \p is the anti-polar plane of p, with reference to an ellipsoid so situated that each vertex of the reference tetraedroii is the anti-pole of the opposite face. If the refer- ence tetraedron is regular, and a be one of its equal edges, the ellipsoid becomes a sphere whose radius is easily found to be 2V2 With this geometric interpretation Pi/> 2 = (172) causes p l to be in the anti-polar plane of p 2 with reference to the reciprocating ellipsoid, and vice versd; PjjP^O (173) causes P 1 to pass through the anti-pole of P 2 , and vice versd; 14^ = (174) causes LI to intersect the anti-polar line of L 2 , and vice versd; P\P=V (175) makes p the anti-pole of P. 64. All the quantities we have to deal with in three-dimen- sional space viz. scalars, points, lines, screws, and planes are expressible in terms of fifteen quantities, which are all either the reference points or products of them of different orders ; they are the four reference points ; their six products, two by two, i.e. the edges of the reference tetraedron ; their four products, three by three, or the faces of the reference tetraedron ; and the product of the four, which is numerical unity. A multiplication table can be easily constructed sim- ilar to that in Art. 45. Considered as an algebra it appears that this system is fifteen-fold. CHAP. II.] MULTIPLICATION. 67 65. Projections. We have the same fundamental formula for projection as in Art. 46, viz. : R (Projection on B of A, directed by C7) = ' ^ , . (176) .LiC in which BC is scalar, while B and C separately are not. If we substitute in the equations of Art. 47 vectors for points, and plane-vectors for point-vectors, we shall obtain a set of corresponding formulae for a vector system in solid space, as follows : p s i + 3-p i 2), (177) s i p| 2 + |ij ' P\es)> - (178) 1 2 3 _ 1 ' P 2 3 _[_ 2 3 ' P l 1 2 3 2 3 1 These are derived from eqs. (101), (102), and (103), and the last one gives p in terms of its projections on ^ parallel to 2*3, and on 2 e 3 parallel to c^ Also, from (104), (105), and (106), we have 1 2 ' 3 4 = e l ' 23 4 -f 2 ' 3 4 1 = 3 ' 4 1 2 ~ 4 ' ^3*3, (180) P = !*2 + 1 3 ' Pi's.' ........ (181) = ' '/W?! + Vs ' if'Mz)) (183) \>wii ylyz), - - (184) T?2*?3 ' Tmi^ (185) (186) 77 = ^ . 77^ + igij ,7713 + tit., 7713, . ....... (187) = ws wi + i i y^ ............ (188) 68 DIRECTIONAL CALCULUS. [ART. 66. Finally, from the equations of Art. 49, we have 1 2 ' l' = 1 ' 2| l' + 2 ' l| l', 1 2 1 l = i| i ' 1721V 2| 2 1 2 1 2 ( 1 2)" 1~ 2 ( li 2) , l! l' l 2 f l| 3 1 2| 2 2 3 (189) (190) (191) (192) (193) (194) (195) (196) 3| 1 C 3| 2 3| 3 P ' ( c l 2)~ == 1 " P f 2\ f l f 2 2 ' P l| l 2- From eq. (192) on, the plane-vector equations have not been written; to obtain them we have only to substitute plane- vectors for vectors in eqs. (193)-(196). 66. Projections in a point system. Write and multiply successively by p\p-2pz, PzPzPn> e tc., and we find PPiPzPs PPzlhPo whence P = x, = -, etc., Oo 'PPiPsPs Pi 'PPsPsPo +P-> 'IWaPnPi Ps'PPoPiPa)) .... (197) which gives p in terms of its projections on any four points. Write next 2 ~~ OJL o ~i ^1? i ~i **^*>?-'*> ?-^3 * X^/A) y i? and multiply successively by Pip 2 p 3 , l^PsPot an d PsPoPi ', the values of x and ^ will be the same as before, but that of x 2 1 will be x = ; hence PoPiP 2 Ps (P'PPiP*P*Pi CHAP. II.] MULTIPLICATION. 69 which gives p as the sum of its projections on p , p^, and the line p*,p s . Similarly, we have for the expression of a point in terms of its projections on any point p t and any plane P 15 ......... (199) and for the expression in terms of its projections on any two lines Z/j and L.,, p== ^'P^ + ^2-p^ ^ (20Q) -/Vji>2 X/jX/o In equations (197) -(200) the projected point p may be replaced by a screw S, or a plane P. We may also write in (197) and (198) planes for points throughout. Let P\=PzPzP^ P-2=PsPopi, etc.; the projection of p on P, directed by \p^ is ' '-Plff^ and we Pipi may write _ \Po-ppo \Pi'P\Pi , P 2 'PP 2 , \Ps-P\P* I P 3 p or, taking the complement of both sides, \P*-P*'pM- (202) Let there be two planes, P = prf^Ps and Q = q^^ ; then PQ = PiPsPa ' ; then PL = LP = q l - q 2 P + q., -q,P, = PI p. 2 p s L -\-p 2 p z piL -\-p 3 the results being found as in previous cases. Multiply the first of (204) into Q ; then qi P q,Q q,P q 2 Q Let Q and R be two planes intersecting in L, and substitute QR for L in (204) ; then, by (205), " _ Pi PiQ PiR I . . (206) Ps PaQ PsR If S be a fourth plane, multiply (206) into it, and we have PQRS = p 2 Q p 2 R pJS (207) Of course three other equal expressions could be written in terms of points in the other planes. In (205) let P= \p l and Q = \p 2) and we have (208) as in eq. (117). In (206) let ^ = 1^ and R = \q 2 ; then PI PI Pi P 2 \

  • 2\<1 Ps\ qa,PM, - (211) in which the second member is a determinant formed on the plan of (210), of which the quantities given make up the first diagonal. In all these equations points may be put for planes and planes for points without affecting their validity. Also, be- cause of the homogeneity of the equations in all the points involved, these points may have any weights we please. 67. Normal form of the screw. Returning now to the sub- ject of Art. 61, we propose to show that by properly choosing the position of the line part of S, the screw can be reduced to a line and a perpendicular plane-vector. The complement as used in treating screws will refer to a unit normal vector system, so that |e will be a plane-vector _L to c and having the same tensor. We have, from Art. 61, S = e 2e + 2Q> - e ) = q$t -(q- e )2e + S(p - e )e. (212) Write, for convenience, 2 = a, g e = p, and 2(i> e )e= \(3 ; .'. S = qa - pa + \fi (213) The condition that the plane-vector |/3 pa shall be perpen- dicular to a is (|/3 /oa)|a = =l/8a pa|a =|a pa \a/3, whence ^ = !^ (214) Comparing the first member of (214) with eq. (176), it appears that it is the orthogonal projection of p, or q e 0} on 72 DIRECTIONAL CALCULUS. [ART. 68. a plane J_ to a. Hence the second member gives the length and direction of this projection in terms of known quantities ; that is, it is the vector perpendicular between the lines e 2e and ?2e. We have by (189) and (214) a(\a pa) a(ap \a) a( a p a + p a-) ~~ ~~~ ~~ /3 Substituting this value of ap in (213), it becomes S ^ (J a + & . \a = <& + 2S g - ^ iSc, . . . (215) and the required reduction is accomplished. 68. Product of two screws. Let the screws be Si = e^f.1 -f- a^! = gjtj -(- ttifi and >S 2 = e 2 2 ~\~ ^2^2 = 6 2 2~l~ a 2i 2 in which a x and a 2 are scalars, called by Ball the pitches of the respective screws. Then Now this is a progressive product, each term being the prod- uct of two lines, and scalar; the two lines in the last term being in the plane at oo, they intersect, and their product is therefore zero. (See also Art. 23.) Further, by eq. (45), = C 1^2 = l2 = zl = 2?7l hence the product becomes ^ 1 /S' 2 = e 1 c 1 e 2 2 + (a 1 + a 2 ) 1 2 ..... . . (216) If $ 2 = $!, we have S 1 2 = 2a ll i. ..'...' ....... (217) If Si reduce to a line, a : must be zero, as appears from the value at the beginning of the article ; hence S 2 =0 ..... : ........ (218) is the condition that a screw shall reduce to a line. CHAP. IT.] MULTIPLICATION. 73 69. The product pS = pee. + apy is evidently a plane through the common line of the planes pee and prj. We wish to show that we have Spi Sp 2 = Spip 2 S apip 2 ...... (219) Take two lines pp' and qq' whose sum is S ; then Sp l Sp 2 = (pp' + qq')p 1 (pp' + qq')p 2 =PP'PI -pp'pa +PP'PI qq'p 2 + qq'pi -pp'p 2 + qq'pi qq'p 2 p' + qq'pip a qq' + pp' -Piqq'pa +p'pi -pqq'pa -p'qq'pa + qq' 'PiPp'pa+q'lh qPP'pz+Pfl q'PP'p'2 = pp'pip a S + qq'pip a S +PI(P- p'qq'p2 -p' -pqq'p* + q q'pp'p* - q' - qpp'p*) = Sp r p 2 S Pip a -pp'qq' S - S - In the above we have used eqs. (197) and (204), and also the fact that pp'qq' is constant whatever the lines may be so long as pp' + qq' = S, for S 2 = 2 pp'qq' = 2a^, by (217). We easily find in the same way, S-pS = ap, .............. (220) Spi - Sp 2 Sp s = a(ih -p 2 p 3 S +p z -pspiS +PZ -pip 2 S), (221) In all these equations planes may be substituted for points. 70. We give here the Quaternion equivalents of some of our vector expressions ; of course there are no such equivalents for point expressions. e^ 3 e 4 = F Fe i*2 ' 34 ' e 5e = S ' FeiC 2 Fe 3 e 4 Fe s e 6 . The superior simplicity of Grassmann's notation is evident at a glance, and the interpretation of the expressions is as much simplified as their form. 74 DIRECTIONAL CALCULUS. [ART. 71. 71. EXERCISES. (1) If rji and rj 2 are two plane-vectors, and PI and P 2 are two point-plane-vectors, show that the bisec- tors of the diedral angles between them are 'T' _l_ 'T' dnrl ~P T"P -4- ~P T~P V)l-lTf)2 ^E Tfjn -L fji clIKl J \ J. 4 2 It JT o -* * 1 respectively. If T?! = cjc^ r; 2 = cie., P l = p^p\p^ and P 2 =^ these become ^i ( ^2 ^1^3 -^ ^> ^i^/ l^ii2^i3P'') *> or, if we write A, ^ -4 2 , ^4 3 for the double J^ areas of the faces of the tetraedron oppo- site p , Pi, etc., the expressions become /<* \^^\ i -A-^l P o (2) Show that p (A l p 1 -4 2 p 2 ^. 3 p 3 ) and A& ^I 2 e 2 A 3 e s are trisectors of the triedral angle at p ; that is, that the first expression is the common line of the bisecting planes through p n pi, Pop*, and p p s , while the second is || to it. (3) The trisector of a triedral angle of a tetraedron pierces the opposite face in a point such that, if it be joined by right lines to the vertices of the tetraedron that are in this face, the triangles thus formed are proportional to the adjacent faces. (4) The bisecting plane through one edge of a tetraedron divides the opposite edge into segments which are proportional to the adjacent faces. (5) The twelve bisecting planes of the diedral angles of a tetraedron pass six by six through eight points which are the centers of the inscribed and escribed spheres. (6) The twelve points in which the edges of a a tetraedron are cut by the bisecting planes of the opposite diedral angles fix eight planes, each of which passes through six of them. (7) Using A , A l} etc., as in exercise (1), show that 1 3 2 2 -4 2 ^4 3 cos < / ' .A. CHAP. TI.] MULTIPLICATION. 75 (8) Show that if e , e } , e. 2 , e are non-coplanar points, and e o> e i> e -2) e 3 divide the lines e e l5 &, e 2 e s , and e 3 e , so that 2V,,' 2V/ 2V 2 ' Te s e 3 ' = Te n 'e l Te/e, . Te 2 'e 3 - Te s 'e n ; then will e n ', e/, e/, e/ be co-planar. (9) If a plane cut the faces of a tetraedron e^e^ in the lines L , LI, L 2 , L 3 , L lying in the face opposite to e , etc., then we shall have the relations (10) By interchanging planes and points derive the recipro- cal propositions to (8) and (9). (11) If two tetraedra e^e^ and e 'e/e 2 'e 3 ' are so related that the right lines through corresponding vertices all meet in one point, then will the corresponding faces cut each other in four coplanar right lines. (12) If n 1} n 2 , etc., are scalars, and LI, L. 2 , etc., lines, and we have the relation n s L 3 + n 4 L 4 = 0, then any straight line that cuts .three of these lines will also cut the fourth, and, consequently, LI, L% L 3 , L^ are generators of the same system of a skew quadric. (13) The perpendiculars from the vertices of a tetraedron on the opposite faces are generators of the same system of a skew quadric. (14) The six planes through the middle points of the edges of a tetraedron _L to the respective edges meet in one point. If i, e 2 , c s are the vector edges of the tetraedron drawn out- ward from e , and p is the vector from e to the common point of the planes, then P = rr (l^s ' r 76 DIRECTIONAL CALCULUS. [ART. 71. (15) The lines joining the corresponding vertices of a tetrae- dron and its complementary tetraedron are generators of the same system of a skew quadric. State the reciprocal propo- sition. (16) The center of gravity of the faces of a tetraedron coincides with the center of the sphere inscribed within the tetraedron formed by joining by right lines the mean points of the faces of the first tetraedron. (IT) There are given six lines Z/ b L 2 , L 3 , e^', e 2 e 2 , e 3 e.,J ; planes pass through L^ L 2 , L s , and cut e^', e 2 e 2 ', e 3 e 3 ' respec- tively, in points which move along these lines uniformly at rates v l} v 2 , v s ; find the locus of the common point of these planes. (18) If P! = P 2 = 2 1 1 , P.. = 2 n e, show that 1 = is the condition that they shall have a common point at oc ; that is, be all || to one line. (19) Show that the condition that Pj and P 2 of the last exercise shall be parallel, or have a common line at oc, is 1 'o m 1 '2 ra 2 m m 2 m 1 1 3 m s = 0. (20) Show that, if P 1 and P 2 be parallel, then IPjPo is a line through the mean point of the reference tetraedron. (21) If any plane be drawn through the middle points of two opposite edges of a tetraedron, it will divide the volume of the tetraedron into two equal parts. (22) Show that p l L } L 2 L^2= p2L s L 2 L 1 p l , and CHAP. II.] MULTIPLICATION. 77 (23) Prove, by eq. (159), the relation (?i 2 + k 2 + ^s 2 ) Oi 2 + -2 2 + wia 2 ) > (Jih + ^2 + Is^sY- (24) Prove, by eq. (193), the formula of spherical trigonome- try cos a = cos b cos c + sin b sin c cos A. (25) Show that i, ]ic, and ie|t are three mutually perpen- dicular vectors, no matter what the directions of i and c may be. (26) Show, by eq. (211), how to express the product of two determinants of the fourth order, as a determinant of the same order. (27) Show that SP = pi -p^PsS +p. 2 -p s piS +p 3 > pip. 2 S, if P = pip. 2 p s ; and, hence, that Sp t Sp 2 Sp 3 = atr S CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 87 CHAPTER III. APPLICATIONS TO PLANE GEOMETRY. 72. In the present chapter, since plane space is under con- sideration, we shall have constantly : The product of two vectors a scalar quantity; The product of three points a scalar quantity. Also, if e , e x , e 2 are reference points, and tj = gj 60, c 2 = e. 2 60, we shall have the relation and furthermore, if p be any unit point at a finite distance, Cl 2 = 1. (222) In this equation of course p^f 2 is a combinatory product of the point and two vectors, and therefore not the same as p times the scalar e^. That p\ (e + i + Cg) = 1 appears also from eq. (105), viz. : p = e p\eo + e^ p\ei + e 2 . p|c which requires the sum of the coefficients of the e's to be unity. We shall have frequent occasion to use the mean point of the reference triangle, and shall designate it by e, so that 3e = eo + ei + e2 , ........ (223) and eq. (222) becomes 3p\e = l ............ (224) 88 DIRECTIONAL CALCULUS. [ART. 73. The equations of curves in plane space may appear under any one of the six following forms : f Expressed in points. Non-scalar equations. < Expressed in lines. ( Expressed in vectors. ( Expressed in points. Scalar equations. -| Expressed in lines. ( Expressed in vectors. 73. The non-scalar equation p = ze + xe v + ye 2 = e + x(e v e ) + y(e. 2 e )> (225) the third member being obtained by the elimination of z, by the aid of the relation x + y + z = 1, which always exists because we use only unit points, may be called the equation of our plane space ; for by giving suitable values to the scalar variables p may be moved to any point of this space. The corresponding scalar equation is 1 2.L^ " ^j ( * ' ) which is simply the condition that p shall lie in the plane e^e.,. If a single condition be given between the scalar variables in eq. (225), such as f(x, y, z) = 0, or f(x, y) = 0, then p will vary according to a fixed law, and will therefore move on some curve. Let L = \p ; then L = 2Je + xfa + y\e z (227) may be any line in the plane e^^, ; but if a relation exist, as above, between z, x, and y, then L will move according to some fixed law, and will envelope a curve. Writing in (225), p e = p, e l e ( , = cj, e., e = e, we have p = xe l + ye,, (228) a vector equation which will represent a curve when a relation exists between a; and y, p being regarded as always drawn out- wards from a fixed origin. CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 89 74. The equations p = z-.... (231) or p\ (ne Q + le + me.,} = 0. If Z = m = w, this equation becomes that of the line at oo, viz.: jpje = 0; ........... (232) for the points ne^ Ie and ne 2 me n , in which the line cuts the reference lines e^ and e e. 2 , are in this case at oo . The equation of a line through any two points, p l and p 2) may be written p = xpi + (1 - X )P2 =P* + *(Pi Pt)> (233) and the corresponding scalar equation, found by multiplying by Pip*, is .......... .:. (234) In general, the equation pL = Q ........... (235) is the scalar point equation of a line, if L be constant and p vary, and the scalar line equation of a point, if p be constant and L vary. Thus, the complementary equations to (231). (232), and (234) are 90 DIRECTIONAL CALCULUS. [ART. 75. L(ne + hi + we.) = 0, ...... (236) ie = 0, ........ ... (237) LL,L 2 =0, .......... (238) which are line equations of the points ne + le t + me. 2 , e, and AAj respectively. If we have such an equation as pp\p z = C, C being a scalar constant, it may always be rendered homogeneous in p; for, by eq. (224), 3p\e = 1, so that we may write PPiPs = 3 Cp\e, or p(piPi 3C- e) = 0, ...... (239) which is a line through the common point of p t p 2 and the line at oo, and is therefore parallel to Pip 2 . For vector equations of right lines we have (^40) T / \ /\ I ' ~ ~ yrf^< f and (p j) e 2 = > the scalar and non-scalar forms of the equation of a line through the end of ej parallel to c 2 . Also ,-,, + <*-,), i and (p ej (c 2 tj) = 0, or p(e. 2 q) = e^, ) for the two forms of the vector equation of a line through the ends of e t and c 2 drawn outwards from the origin. The equation P e = C (242) is that of some line parallel to e ; while P U = <7. . (243) is that of some line perpendicular to e, as is easily seen from the meanings assigned to pe and p\e. in Chap. II. 75. Transformation of scalar equations from a point system to a vector system, and vice versa. Take e for the origin of vectors, and write p e = p, 2h e o = ti> e tc., the difference CHAP. TIL] APPLICATIONS TO PLAXE GEOMETRY. 91 between each fixed point and the origin being equal to some constant vector. Thus to transform (234) to a vector system, we have = Ou + p) Oo + i) Oo + ft) = <\,Oi + ci 2 + *2p) = 0. The term p^e, vanishes by Art. 21, being the planimetric product of three vectors ; also e^p^ = pe l} etc., hence the equa- tion becomes p(e 2 e i) = *i c 2> * ne same as (241). Since in changing from a point to a vector system we have dropped the point e from each term, it follows that in the reverse change we must first multiply each term by some fixed point. Thus to change (p tj) e 2 = to a point equation we have (p - i)2 = e *(P -Pi} (Pi - e o) = e (p Pi)p a = 0- 76. EXERCISES. (1) Find the equations of right lines satisfying the following conditions : Passing through p l and parallel to LI ; Passing through 2h and parallel to e ; Passing through the common point of two right lines and having a given direction ; Passing through the common point of two right lines, and also through the common point of two other right lines ; Passing through the end of ^ perpendicular to e,- Ans. (p p^Li = 0, pp^ = 0, pe.L^L-2 = 0, (2) Interpret the equations obtained by putting lines for points, and points for lines, in the first four results of the pre- vious exercise. (3) Find the common points of the following pairs of right lines, (4) Find the condition that the three lines p\p\=, |p 3 =0 shall have a common point; also the lines P|CI= C^ 2 = C p[f 3 = 3. Ans. prfzps = 0, and C^s + CWi -f C&c z = 0. 92 DIRECTIONAL CALCULUS. [ART. 77. (5) Show that the common point of the two lines ppip 2 = C 3 , pp 2 p s = d is p 2 + - - [<7 3 (>3 - Pi) + 3 ) ]. (6) Show that, if Pip 2 p s = GI + C 2 + G' 3 , then the three lines ppip 2 = C 3 , pp 2 p s = Ci, PPsPi = C 2 have a common point. (7) Show that if the equations of three lines, on being mul- tiplied by any constants and added, vanish identically, that is, for all values of p or p, then the lines have a common point. Show also that the results in exercises (4) and (6) are in accordance with this. (8) Find the condition that the three points whose line equations are L\Li = 0, L\L 2 = 0, L\ L 3 = shall be collinear. (9) Show that the perpendiculars from the point e on the lines whose equations are pL l = 0, pL^ = C, (p Pz)p\p 2 = 0, are respectively of the length ^, eL i~ c t PiP*( e ~P*\ (10) Find the vector perpendiculars from the origin on the lines ep=C and e\p = C. Also from the end of e' on the same lines. C , Ce (11) If L l =p 1 c 1} L 2 =p 2 e 2 , etc., show that We have, by Art. 61, since the lines are all in one plane, 2L =(e + a) 2e ; but e + a is a unit point, hence T^L = TSe. (12) Show that T(L + BC\e)= TL. 77. If LI and L 2 are two straight lines, then the equation L v p L 2 p '= represents the two lines simultaneously, for it is satisfied whenever p lies on either of the lines. The equation L 1 p-L i p = C ......... (244) represents a locus that evidently differs less from being the two lines L t and L 2 , the smaller C is ; also, when p is indefi- nitely far from L 19 it is indefinitely near to L& and vice versd. The locus is of the second order ; i.e. it is cut in two points by CHAP. IIL] APPLICATIONS TO PLANE GEOMETRY. 93 a right line ; for let p = e + a* be the equation of some line j then, substituting this value of p in (244), we have a quadratic in x determining two points in which the line cuts the curve. The locus must therefore be a hyperbola. If C be positive, Lip and L 2 p must have like signs ; hence p must be on the same side of LI and of L& i.e. in the exterior angle, while, if C be negative, p must be in the interior angle. Thus for the same numerical value opposite signs of C correspond to a primary and conjugate hyperbola. The complementary equation p l L-p,L = C (245) represents the reciprocal curve to (244) . When (7 = 0, it represents the two points p l and p 2 and their connecting line ; for it is satisfied when L passes through p l or _p a or through both. When C is positive, p^L and p 2 L must have like signs, and hence L must not pass between p 1 and p. 2 ; if C be negative, L must always pass between p l and p 2 . The curve enveloped by L is of the second class, i.e. two tangents can be drawn from any point ; it is therefore a conic. 78. It is easily seen, as in the last article, that the order of the curve represented by any scalar equation in terms of p, i.e. the number of points in which it can be cut by a right line, is simply the degree in p of the term of highest degree in the equation. The equation ALtf) - L s p + BL 3 p . Lip + GL^p L 2 p = 0, . (246) in which A, B, C are scalar coefficients, represents a curve of the second order passing through the points L^L-y, L 2 L 3 , -L 3 A ; for each term is of the second degree in p, and the equation is satisfied when p is on any two of the lines simultaneously. The complementary equation ApzL - p 3 L + Bp s L p^ + CfrL p 2 L = . . (247) causes L to envelop a curve of the second class tangent to the three lines j9,p 2 > P*Ps, PsPi > f r ^ is satisfied when L passes 94 DIRECTIONAL CALCULUS. [ART. 79. through any two of these points simultaneously. As an exer- cise let the student interpret the following equations, k being a scalar constant : PiPzP ' PsP*P = piPiP - PtPaPi 'PsPtP PiPtP s PiP*PfPaPsPi LJ^LJ^LJ x/3-L/4.Zv = I\,LJ\LJ^LJ IJ^ -LJlJ-J-2-Lj ' XVgi/^iV JL/jJL/^i/ I.I.I: 79. Differentiation. Before proceeding to the general treat- ment of equations of the second degree in p and p, we will consider the question of differentiation as applied in this cal- culus. Let p = ze + xe + ye 2 = e c + x(e l e ) + y(e 2 e ) = e + ct-t! -f ye which implies that x + y -f- z = 1, as we always assume. If p move from point to point, it is a function of the time, as are also x and y ; hence ^ = l ^ + C2 ^. (248) dt dt 2 dt Thus the differential coefficient of a point is a vector. Also, since p=p e , 3e = ^. . (249) dt dt If Tfi = Te 2 = 1, ei|c 2 = 0, and a relation subsist between x and y such as f(x } y) = 0, so that p moves along some curve, /dY / dy\* , fdy\ 2 fds\ 2 then [ -^- = [ e, + e,-- ) = 1 +/ 1 =1 ) , \dx) \ l 2 dxJ \dxj \clxj' T \dxj dx whence T^ = l, ........... (251) CHAP. TIL] APPLICATIONS TO PLANE GEOMETRY. 95 Let L = \p = e + xe. 2 (e + e^) y0i(e 2 + e ), so that L is sub- jected to the same condition, z +- x + .V = 1> that j9 is, which, however, affects only its length, and not in any way its position ; dL ( \ \ dy , , \ dx J dx then (252) dL Multiply - - by e, and we have e = dx dx = ; (253) dL hence -- is a line through the mean point of the reference dx triangle. By the figure it is evident that pp' or p p' is a chord of the curve which is the locus of p ; as P' approaches p, p p' approaches the tangent at p in direction, and at the limit has this direction ; hence limit of *>-*>' = ^L = ^> T(p-p') Tdp ds is a unit vector along the tangent at >. Similarly, if L envelops some curve, dL TdL (254) is the limit of as L' approaches L. But L L' is always a line L-L' T(L - L'} through the common point of L and L', which ultimately becomes the point of contact of L with the curve. Hence is a unit line through the point of contact of L and the dL TdL mean point e. If a scalar equation in p, L, or p be differentiated, it will necessarily become a homogeneous, linear function of dp, dL, or dp, and thus independent of the length of dp, dL, or dp ; we 96 DIRECTIONAL CALCULUS. [ART. 80. may therefore, if we please, regard these not as infinitesimals, but as finite in length. Take for instance the equation P\ e o ' P\ e i -\-pL\ + pL 2 - pL s pL 4 = ; differentiating, we have dpe -pe l + p\e dp e^ + dp^ -f- dpL 2 pL s pL 4 + pL 2 dpL s pL 4 + pL 2 pL 3 - dpL 4 = ; and, as dp appears once, and only once, in each term, it is evi- dent that its length may be taken as great or as small as we please without affecting in any way the meaning of the equa- tion. 80. Examples of differentiation. As shown by the example just given, the process of differentiation does not differ in principle from that of ordinary algebraic equations ; we have only to pay attention to the alternative law of multiplication. d(pp l L 1 p 2 p') = + prfp pp^ V . . (255) d(peq)= dpeq -i-pedq = e(qdp pdq). . . . (256) .. .' . . (257) (258) dT 2 P = 2TpdT P = d(p*) = 2 P \d P . (259) = U P 'Up\dp+TpdUp ) Also, by eqs. (189) and (260), . . . . (261) The student will find it interesting to examine the geomet- rical significance of the last three equations. CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 97 81. Tangent and normal. If the equation of a curve be given in the form p = <(z) = a*, +/(*) . eg, ...... (262) then, as -^ is a vector along the tangent at the end of p, if we Q9P let )- - 2 2 is that of some curve passing through the common points of the two circles. The first member of the equation is the square pf the distance from the end of p to the point of con- tact of a tangent to the first circle drawn through the end of p, while the second member has a corresponding meaning for the other circle : hence the equation is the locus of points from which equal tangents can be drawn to the two circles. Expanding, it reduces to ..... (271) a straight line called the axis radical. 83. EXERCISES. (1) Show that the equations and z = k' represent circles, and find their radii, and the vectors to their centers. Also, if C= C" = 0, show that the two circles cut each other orthogonally, (a is some constant vector.) (2) Show that the three axes radical of three circles have a common point. CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 99 (3) Show that if e^ c 2 , 3 are three vectors drawn outward from a common point, and they are connected by the relation 1 2 ' r + 23 ' l ? + 3 1 ' 2 2 = 0> then their outer ends lie on a circle through their common point. (4) By eq. (97) and the relation given in the last exercise show that the equations r r 2 2 , 2 ~ 3 i ,2-2 3- r 1 2 2 3 3 1 also hold between three vectors which, being drawn outward from a point, terminate in a circle passing through this point. (5) If perpendiculars be drawn from a point upon the three sides of a triangle, and the feet of these perpendiculars be col- linear, then will the locus of the point be a circle circumscribed about the triangle. (6) Show that the tangent line to the circle of eq. (267) has the equation o- = a(ij cos + i 2 sin 0) + wa(i 2 cos ii sin 0) , of which the scalar form is o-|p = a 2 . Also the equations of the tangent and normal to (269) are respectively ( dy 2 a a vector parallel to the tangent at the end of p\ hence the equation of the tangent may be written 1 + A ..... (274) in which y is to be taken as constant. Eliminating y from (273), we have the scalar form of the equation, viz. : (p| l2 ) 2 = 4a.p| tl , ......... (275) or, as it may be written, In this latter form we see that the vector i 2 p|i 2 4 ai t is always perpendicular to p. Let a- = t 2 p t 2 4 aii ; then |i 2 = p]t 2 p\ti + 2 a (p 0-) |t2. (4) Show that p = ^t\ + fc 2 and (cjp) 2 + 2 C& *& 0, are respectively the vector and scalar forms of the equation of a parabola referred to a tangent whose direction is e 2 and a 102 DIRECTIONAL CALCULUS. [Aitx. 86. diameter through the point of contact whose direction is that of ej. (5) Show that, with reference to the equations of Ex. 4, the lines a = p 4- x-P and a = p + yt 2 cut the line o- = z^ at equal at distances on each side of the origin, and give the geometric interpretation. 86. Ellipse and hyperbola. The equations p = aii cos 6 + bi 2 sin ....... (276) p = aii sec 9 + bi 2 tan 6 ....... (277) represent respectively an ellipse and hyperbola, in which a and & are the semi-axes, and is the eccentric angle. This will be evident at once to any one familiar with the ordinary Cartesian equations, if we obtain the corresponding scalar forms. We find p\ii = a cos 0, p\i z = b sin 0, from (276), whence 1 ......... <278) From (277) we have p\^ = a sec 0, p|i 2 = b tan 6, whence ' ......... (279) Since p|ij and /o|i 2 are the Cartesian x and y, the equations are evidently those of the ellipse and hyperbola. The two equa- tions may be combined by using the double sign ; thus EXERCISES. Show that the equations T(p + c tl ) = - 2 + P \ and T( P - c' tl ) = ? in which c = Va 2 b 2 and c' = Va 2 + 6 2 are equivalent to (278) and (279) respectively, and interpret these forms of the equations. CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 103 Show that the equations r(p + ci,) + T(p-a l ) = 2a and T(p + c' tl ) - T(p c'O = 2 a are also equivalent to (278) and (279) respectively, and inter- pret them. Let us write the equation ts* ........ (281) This expression is a linear and vector function of p, and, by the aid of (281), equation (280) becomes P \P = 1 ............ (282) This remarkably simple equation may represent, as will appear hereafter, not merely (280), but any equation of the second degree in p, which contains no first-degree terms. It may thus represent not only any central conic with the origin at the center, but also any central quadric referred to its center as origin, when we are dealing with three-dimensional space, and the corresponding locus in n-dimensional space. Similarly, if p be a linear, point function of a variable point p, we shall see that p\p = may represent any conic whatever in two- dimensional space, any quadric whatever in three-dimensional space, and any locus of the second order in 7i-dimensional space. In the form given above the function will be found to possess the following properties, viz. : (a) (y) (8) The first three properties are possessed by any linear, vector function. When the last relation holds, the function is said to be self -conjugate, and, in dealing with curves and surfaces of the second order,

    p 4- p\$dp = 2dp\p = 0. Hence

    p = 0, or, by (282), o-i^ = l, ........... (284) as the equation of the tangent to the curve. For the normal we have the equation (o-- P ) = .......... (285) Since is parallel to the normal at the end of p, the projec- tion of p on tf>p will be the perpendicular from the center on the tangent line. By Art. 46 this is ftp . p\p P 1 -&?F m tiff- r*- u * (286) Hence the length of the vector p is the reciprocal of that of the perpendicular from the center on the tangent. 88. Diameter. The diameter being the locus of the middle points of a system of parallel chords, we may find its equation as follows. Let the system of chords be parallel to e, and let p = p l + xe be the equation of one of them, in which p l is a vector of the curve, i.e. pi\pi = 1. Substitute this value of p in (282), and we have, in order to find the other end of the chord, = , or whence, because of the condition above, x = Q, and x = -- TT-^- CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 105 Now, if or is the vector to the middle point of the chord, we have Multiply into [ function. Write p = 9i l i p|'i + Ste p\h) ..... (289) so that, comparing with (281), we have 0! = ^ 02= | ........ (290) Putting, in (289), successively i t and i 2 for p, we find <'i = 0i*i and p for p in (289) ; therefore = tfp = 0i l i P\ l i + 9& - P\ 1 2 9l 2 <-l ' P\ L l + 02% ' P\ 1 3' 106 DIRECTIONAL CALCULUS. [ART. 89. We have the fourth member from the third because by (79) and (283). Similarly, < (< 2 p) = * P = g^ p\ Ll + g^ . p\i^ etc. ; so that, if n is a positive whole number, = S^i P!*I + 0sV /12 ...... (292) Let ra be some other positive whole number ; then also <> = 9\ M >-i P\ l \ + d2 m h p\* ; hence n

    n+m p = 9"<-i i-i\ m p + 92 n h <-2\ n p -0r + %.A + fc" +I Vpi*2. - . (293) Suppose m to be negative and equal to n ; then by (293) <" (<" = 4>p = l i ' P l i + '2 PJ4 = p, so that the negative exponent gives a function, such that the operation indicated by with a positive exponent of the same numerical value being performed upon it, gives p as a result ; i.e. the operations < n and <~ n cancel each other. Hence (292) and (293) hold both for positive and negative values of the exponents. Finally, suppose n = - ; then we ought to have if the exponential law holds for this case, and the result is easily verified as before. Thus (292) holds for all real values of n. If /() = An? + Bx n ~ l + . . . N, we may easily show that * 2 ' Pi 1 * (294) in which 4> =/(^) is also a linear, vector, self-conjugate func- tion. CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 107 Again, if f^ and / 2 are functional symbols of the same form as /above, let * =yTT = ^W 1 5 then J-2(9) -^ (296) so that * is still a function of the same kind as <. Finally, let <, <', <" be three functions of the form (289) with corresponding g gj, gj', g* g 2 ', g 2 " ; then k P + k'& = (kg l +k'g l '+k"g l ") ll . p | 4l . (296) All the results of this article hold equally well for n-dimen- sional space when tf>p = 2"(g < p|t)- 90. EXERCISES. (1) Show that Cl \O/ (2) Show that ^( 2 -l)-'i 1 = J. ^~ Ot (8) Show that (4) Show that (5) Show that, if 'p = (}>'p. (6) Show that 108 DIRECTIONAL CALCULUS. [ART. 01. 91. The function = ^fy* we have p\p = P |**(**p) = **p|** P = (^3=<*/3|<^a; (298) hence ^a and ~(3 are unit normal vectors. We will determine what relation (f>*p bears to p in the ellipse. and p'\'p' = afi^p 1 = p', and p'^p = 0, so that and since they pass through the end of c, the equations must be satisfied when e is substituted for cr. Thus we have the relations li = 1 = |^p s , which must always hold between c, p l} and p 2 . If now, in the given equation o-|-'-k a )- 1 p = l, and (

    (p 1 -|-p2) = 0, and that these diagonals are conjugate in direction. (5) Find the condition that the line cr|e = C shall be tangent to the curve p|

    p = 1 ; comparing this with the given equation, we have p = , or p = ^ -. Sub- G C stituting this value of p in the equation of the curve, we have y I = 1, or C f2 = e|^)~ 1 c, the required condition. Thus the C line whose equation is r^>- e ...... (300) is always tangent to the central conic. 94. TJie conic referred to conjugate diameters. The equation ( a pr ...... (301) 112 DIRECTIONAL CALCULUS. [ART. 95. represents an ellipse or hyperbola referred to the conjugate semi-diameters a and ft. The equation is satisfied when p = (3, so that the curve passes through the end of ft ; also, the curve is satisfied when p = aVl, so that the end of a is a real point of the ellipse, and an imaginary point of the hyperbola. -ITT -j. la ap , \B Bp Wnte *_L_J| Jfe ( so that (301) becomes pjp = 1 ; then we have and a and ft are conjugate in direction, by (288). Equation (301) shows that, in the hyperbola, if any diameter cuts the curve, its conjugate does not. EXERCISE. Show that the ellipse and hyperbola of (301) are also represented by p = a cos + ft sin 6 and p = a sec 6 + ft tan 0, respectively. 95. The area of the parallelogram formed by tangents at the ends of conjugate diameters of an ellipse is constant. If a and ft are any pair of conjugate vector semi-diameters, then the area is 4 aft. Now a = ~%(f>*a = CU, ijl^a 4. fo, . t^^a, hence - = 4 ab - tl is <>?a<> = 4 ab. See (298) and (193). 96. EXERCISES. (1) Find the locus of the intersection of tangents at the ends of conjugate diameters. Taking a and ft as usual, we have o- = a + /3 = vector of point whose locus is required. Therefore CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 113 a similar curve, which is, in this case, an ellipse. If the given curve be an hyperbola, the locus of o-=0, the asymptotes of the curve, because, in this case, if /?!p ; also of n p, not- ing particularly the case when n = . (3) Show that, when T<^p = c, the locus of p is the curve whose equation is p\ 2 p = c 2 . (4) Show that the locus of the foot of the perpendicular drawn from any point to the tangent to the conic, i.e. the pedal curve, has the equation (o- e)\<]>~ l (~ l c 2 = a? 6 2 . The conditions give c 2 = c i- Thus we have and eal^' 1 ^ = M^'Vi = 2 (iii) 2 whence, adding, we have the result. (6) If a and ft are vector conjugate semi-diameters, show, by the result of the last example, that a- + f& = a? ft 2 . By Art. 91 <^a and -(3 are unit normal vectors ; real, for the ellipse, and imaginary, for the hyperbola. Hence -* = 2 6 2 = a? (7) Show that the locus of the common point of perpendic- ular tangents is a 3 = a 2 b 2 . Use eq. (300). 97. TJie general linear, vector function in plane space. The most general form of this function may be written ^P = c 1 .c 1 'jp + e 2 .e 2 '|p: ' ...... (302) for, suppose there were other terms such as e 3 e 3 ' p + kp ; write 3 ' = nify' + wW and p = - (|e,' e 2 ' p |e 2 ' e/|p), 114 DIRECTIONAL CALCULUS. [ART. 98. and these terms become k \ f k ' r / k \ f (mi 3 -- . e 2 ' ) p e! r -f ( m& V i 2 J \ which, on being combined with the terms composing (302), give an expression of the same form as before, merely having instead of ^ and e, the vectors -- eg' and e 2 + ??i 2 e 3 + - r - l je/. e l 2 1 2 This general form of < possesses all the properties given in (283) except the last, or self-conjugate property. If, how- ever, we write ......... (303) then we shall have p\(f> c p and p|p ....... (304) c is called the conjugate function to , and vice versa. When c = (f), the function is self-conjugate, and the condition (8) of (283) is satisfied. In all geometrical applications it is possible to take < self -con jugate, and we shall always so regard it, because it greatly facilitates necessary operations. The sum of a linear function and its conjugate is always self- conjugate : for c cr 98. Inversion of the function. Suppose we wish to find the value of p from the equation p = e .......... (305) This gives P = <~ I ; ........ (306) and if we can find the form of the inverse function, we have the solution of (305). For convenience write e = |A, so that p = \\ and p = ~ l \\ ; then whence p = x\ c X = CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 115 To determine x, let /u. be any vector whatever ; then = /xjjA = A/A. and p = 4,- 1 * = -i\\ = (307) This equation affords a solution of (305), but we proceed to obtain a formula more convenient than this for most purposes. Let A and /A be now any two constant vectors whatever ; we have, from (305), \\p = pj< A = Ajc and /A|

    c p. p. ; also, writing p in terms of its projections on \cf> c \ and |< C /A, we have P = TTT or, substituting from the preceding equations, ' terfi.Xic), . (308) the proposed formula. If < be self-conjugate, then <^> c = <, and the suffix may be dropped in (307) and (308). To invert such a function as fa + k 2 + k'fa + etc. = $, say, we have 4> c = c^>i c + ^<^ 2c + k' 3c + etc., and $ and $ c are to be substituted for < and < c in (308). For example, suppose $ = < -f- g ; then + V + A/A if we write 1 = . . . (309) A/x A/A 116 DIRECTIONAL CALCULUS. [ART. 98. Hence (m + m# + (f)p = = g P 4- (h /Tim or + f> = iT" Thus we have still another inversion formula, which is, how- ever, not generally so convenient as (308). Operating by <, (310) assumes the form (311) The quantities m and m x are invariants, i.e. their value will be unchanged if other vectors be substituted for X and p.. For, let #jX + 2/i/i and x^. + y^. be any other two vectors in the plane space under consideration, and substitute them for \ and , : A/i Similarly, (*iX + a^OV V From the above it follows that, in any given case, we may assume such values of X and /x, as may be most convenient. For example, take p = tj ei\p + e 2 e 2 'ip> ^ i n (302) ; let X = | a and A = |c 2 then (i 2 ) 2 CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 117 99. EXERCISES. (1) Invert $? = 7^or> (a alp + (3\p). ( a p)~ Find p as a function of c in the following cases : (2) a.ap + /?./?|p = c. (3) Ap + a-a\p = f, A being a scalar constant. (4) Ap+a. 100. The general equation of the second degree in plane space. This equation may be written y\p=C. ......... (312) For every term of the second degree in p may be expressed in the form p|e p|e f , so that the portion of the left-hand mem- ber which is of the second degree will consist of the sum of a series of such terms, i.e. 2" (pje p|e') . This form evidently includes such forms as (p|c) 2 , equivalent to p|c-p|e, and Afi equivalent to A((p\i l } 2 + (p\iv) 2 + (pl^Y) If we write now 2" (e p e') = p, but will not be, in general, self-conjugate. If, however, we write iS:(e.pje' + e'.pj)=*p, ...... (313) then the second-degree terms will still assume the form and will be self-conjugate because it is the sum of a linear vector function and its conjugate. See Art. 97. The self -con jugate form of < for any set of second-degree terms may be obtained by differentiating these terms, a method which may be convenient for the beginner. For suppose not self-conjugate in the expression pl

    p + p\dp = dp\p + dpl +

    + c is self -con jugate. Also, P\$cp = p\p, so that 118 DIRECTIONAL CALCULUS. [ART. 101. For example, let p\p = p" -f pe pie'. Then, 2dp\p+dpc p!t'+pc -dpje' =dp\(2p e-pe' + e'-pe) is the differential ; and if we put (ftp = p -f- (e' pe je p|e'), we have p|

    /u, differs from zero, the locus has a finite center ; but when \fji = 0, the center is at oo, unless the numerator is also zero. In this last case, 8 becomes indeterminate ; it is in fact the vector radius to a straight line, the locus of centers, whose equation is and (312) now represents two parallel or coincident right lines. CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 119 The equation .......... (316) requires that the function should consist of a single constant vector multiplied by a scalar factor, or should be reducible to that form. For, taking the general form of < as given in Art. 97, we have = (e! Xje/ + e. 2 X c 2 ') (e t - /t^' + c 2 //, e 2 ') Since X and /u, are to have any values we choose to assign, this equation can only be satisfied by making Cjc 2 = 0, i.e. 2 = nti. Thus we have = e! P ; (/ + ne 2 '). But we are dealing only with self-conjugate functions ; therefore the vector appearing in the scalar factor must be the same as the other; i.e. we must have the function of the form Hence, when (316) is satisfied, and the numerator of the value of 8 is not zero, eq. (312) takes the form (e|p) 2 + 2X(p=a ........ (318) Since we know now that the center is at oo, change the ori- gin to a point on the curve, by putting p' + 8' for p. .-. (p=C-8\8-2y\S=C + y\- l y = C', say. . . (320) Let us find the points at which the tangent is perpendicular to p ; i.e. p is parallel to p, or ..... (321) g being a scalar constant to be determined. Eqs. (320) and (321) give ri*-0-.iAflrfP-TV-l . . (322) 7 Multiply the complement of (321) successively by any two vectors X and /u. ; therefore and In order that p may be simultaneously perpendicular to ( g)X and (< g)p, these vectors must be parallel ; hence we must have or g* wiigr + m* = 0, ....... (323) m and raj having the values given in (309) . Eqs. (322) and (323) show that Tp has two pairs of values, the values in each pair being numerically equal but of opposite sign. Thus there are four points at the opposite ends of two diameters at which pp = 0. Let gi and g 2 be the two roots of (323), and p^ and p 2 the cor- responding values of p. Then, by (321), we must have pi = 9i The most general form of a self-conjugate function may be written p *2 CHAP. III.] APPLICATIONS TO PLANE GEOMETRY. 121 whence, by tke conditions above, 2 pi = gi(fi pile/ + c/ PI|I) + g s (fs Pi\e2 + ft and 2p 2 =g l (e l . p 2 |i' + e/ p 2 i/ + 02(^2 pzh' 4- 2 r Hence p^' = pi c 2 = P2\ f i Pz\ i = > therefore c 2 an( i *%' & re both perpendicular to p 1} and hence parallel to each other, and t! and e/ are both perpendicular to p 2 , and hence parallel to each other. Therefore let e/ = ej and c 2 ' = c 2 j then we have Hence cj is || to p l and c 2 is || to p 2 , also cj ? = 7 72 e 1 =l = e 2 i = T 2 e 2 ; thus Ci and c 2 are wniY normal vectors, and it appears that, whatever may be the original form of 0, it may always be reduced to the form p = y\ L i ' p <-i i x and i 2 being unit normal vectors parallel to the values of p which satisfy the equation (321). Now, by (322), C' C' C' C' (/! = = , say, and g 2 = = , say ; r 1 r 2 so that has the form of (281), and (320) represents an ellipse or hyperbola, according as g l and g 2 are both positive, or one is positive and one negative, provided that C' be a posi- tive quantity as it may always be taken. If g and g 2 are both negative, the curve is imaginary. The vectors ( #i)A. and (

    p = Ap + we p e ; suppose that Tt = 1, and put A = e and fjt. = e ; then 00 A. = (<^> -4)e =7ie is _L to the a axis, so that the axes are now completely determined. CHAP. III.] APPLICATIONS TO PLANE GEOMETJRY. 123 (2) Discuss the following equations : (a) (ap)'(#,) 2 = (a0) S . (6) p*-a|p.0|,o--ap.j3p-(a+/8)[p=l, when Ta=T(3=l. (C) (pj(a-/3)) 2 -(pj(a + /?)) 2 = 4(a/?) 2 , When Ta=T(3=l. (d) t ? p ? = e 2 [cj(p e)] 2 , e being a scalar constant. In the last equation consider the cases when e>l, e~ l y =* is the condition that (312) shall represent two straight lines, real or imaginary. (4) Find in how many ways a conic passing through the four common points of two given conies can be reduced to two right lines, using the condition of Exercise 3. (5) Show that, if two conies have their axes parallel, any conic passing through the common points of these two will have its axes parallel to theirs. (6) Hence show that, if a pair of right lines be drawn through the four common points of a circle and any coaic, the bisectors of the angles between these two lines will be parallel to the axes of the conic. CHAP. IV.] SCALAR POINT EQUATIONS. 133 CHAPTER IV. SCALAR POINT EQUATIONS OF THE SECOND DEGREE IN PLANE SPACE. 104. We need only consider homogeneous equations ; for, by eq. (224), ope = 1, so that a term of any degree in p may be raised to any other degree by multiplying by the proper power of 3p\e. Any homogeneous equation of the second degree may be written in the form Ptop = 0, ....... (325) in which is a linear self-conjugate function ; for the left-hand member of such an equation can always be reduced to the sum of such terms as Aipfa p\qi ; that is, to the form l(Ap\q .p\q') = &\-p\(Aq .p\q' + Aq' if we write 4>p = &lA(q.p\q' + q'.pq)l ....... (326) Of course we may have q = q' for some terms of the summa- tion. 105. Eq. (325) represents a curve of the second order ; that is, it is cut in two points by any right line. For, let p=xq l -{-yq a be the equation of some right line, and substitute in (325) : whence y = ~ x q. 2 q. . 2 V- i 2 i8_ (327) q 2 \q 2 134 DIRECTIONAL CALCULUS. [ART. 106. As ^ has two values, it appears that the line must cut the v curve at two points. If the two values of ^ are equal, the line must be tangent to QC the curve. The condition for this is gig 2 | = 0; ." ........ (328) and when this condition is satisfied. - = -- ^ - : so that the x q 2 \q., equation _ gi ' gaga ~ 31 . 106. Diameters. To find the locus of the middle points of a system of parallel chords of the locus of (325). In (331) let q l be on the curve, so that we have q^qi = ; then y = iilz!5. At the middle point of a chord having the direction , we have p = q l + \ ye = ^ 1 c. (332) CHAP. IV.] SCALAR POINT EQUATIONS. 135 This is the equation of a diameter conjugate in direction to e. Let PI and p 2 be any two points in this diameter, so that we have p l \<}>t = Q and p 2 |p passes through p\ consequently \p is the tan- gent line to the locus at p. The equation of the tangent line is therefore tf|*P = 0, ........... (334) q being a variable point, and p a point on the curve. If e be some point not on the curve, let us determine what line \2h \$p 2 . But as they pass through e, we must have and e\(f>p 2 = Q These conditions show that \$e passes through p^ and p 2 , the points of contact of the tangents drawn to the curve from e. e is therefore the polar of e. Let q be any point on |e = = e\q; so that, wherever q be situated on the polar of e, its polar always passes through e. Thus if a point move along a straight line, its polar passes through a fixed point, the pole of this line ; and, reciprocally, if a revolving line pass through a fixed point, its pole moves along a fixed line, the polar of this point. Equation (332) shows that a diameter is the polar of a point at oo. Hence the polars of all points on a diameter have a common point at oo ; i.e. they are parallel to the diameter con- jugate to this. 136 DIRECTIONAL CALCULUS. [ART. 108. 108. Center of the locus. The center is at the intersection of any two diameters ; hence q c = m^e^e., .......... (335) is the center, m being a scalar factor so taken as to make q c a unit point. To evaluate m, multiply both sides of the equation into 3\e ; therefore 3q e e = 1 = 3m\<}>e l f 2 e = 3 met l e. 2 so that (335) becomes If we have efafa = 0, ........ . . (337) while jtfi = ..... (339) is a set of conjugate points. These equations cause each point to be on the polar of each of the others ; that is, the points are the vertices of a self-conjugate triangle, in which each side is the polar of the opposite vertex with reference to the curve There is an infinite number of such sets of points ; for take any point in the plane of the curve as q l} then any point in the polar of q l as q.^ whose polar will pass through q l} by Art. 107, and will cut the polar of qi in q., t . If one point, say q s , is at QC, the other two will be on a diameter ; if q. 2 be also at oo, then q l will be at the intersection of two diameters, that is, it will be at the center; thus q c , ej, e 2 form a conjugate system if we have q c \fi = i|q a an( i we see that conjugate directions are only a particular case of conjugate points. CHAP. IV.] SCALAR POINT EQUATIONS. 137 110. Normal system of conjugate points. If a system of conjugate points, besides the conditions (339), satisfy also the conditions ?s!9i = 0, ...... (340) they may be called a normal system. We proceed to show, that with reference to any curve represented by the equation p\p = 0, there is one, and only one, normal system of conju- gate points. 111. Solution of the equation pp = 0. This equation is equivalent to p = np, or ( n)p = 0. Multiply the com- plement of the first member by any three points K*-)fe0 I ..... (341) %!(* - n)P =p\( - n)q, = J Each of these equations must be satisfied by the same values of p which satisfy the given equation ; i.e. they are simulta- neous equations. The point p must therefore be simultaneously in each of the three lines |(< n)g u |(< n)q^ [(< n)q 3j which requires that these lines shall have a common point, the condition for which is H ' qs . (342) or ?r K 2 n 2 + k^n k = in which Jc = fyq-^q^q* -f- q\q^ls ' . (343) The fc's are invariants; i.e. they have the same values what- ever position the points <7 b q^ q z may occupy, which may be shown as in Art. 98 in the case of m and m^ The solution of (342) will give three values of n, which, substituted in (341), will give the required points at which p = np. Let the roots 138 DIRECTIONAL CALCULUS. [ART. 112. of (342) be n t , n. 2 , n 3 , and let the corresponding values of p be PD P& Pz > then, by (341), the equations Pi\( ni)q 1 ( n l )q 2 = "1 p 2 \( n 2 )q l (<}> n 2 )q 2 = V .... (344) Ps\( n)qi( %)^2 = J give the points p^ p.* p s . 112. To show that the points just determined form a nor- mal conjugate system. We must have <^p l = n^, p. 2 = n.^, p 3 = n s p s . Now, as the function < is self-conjugate, write it in the most general form of such a function in terms of the p's ; that is, 2<&> = nifa -p 2 p s p +\p 2 p 3 -Pi\p) + n 2 (p 2 -Pzpip + \p 3 pi -p t \p) + n 3 (p 3 -pipsp + \PiPz -p 3 \p). If this value of satisfies the conditions above, we must have P!\PI = p 2 \p s = pslp! = and p^ps = p = p} J)/ = 1 These conditions give at once 2 <&>! = nfa -{-\p2ps), etc. ; but they also cause p l to be on the lines \p 2 and \p & simulta- neously, so that Pi = m]p 2 p 3 , m being some scalar constant. Hence Pi\p\ = mpip^pzi or m = 1, so that the required condi- tion p t =. H^P! is satisfied ; and so for the others. Finally Pi\4>p2 = n 2Pi\P2 = 0, etc., so that all the conditions of (339) and (340) are satisfied, and hence the points p^ p.^ p s form a normal conjugate system. It appears then that, what- ever be the original form of , one set of three points may always be found such that, if these be taken for reference points, (f> is reduced to the form P = WiPi -p\Pi + n 2 p 2 .p\p t + n 3 p 3 .p\p s . . (345) Note that these will not in general be all unit points ; for if we express p^, p& p s in terms of the original reference points, we have nine constants, and have subjected the points to seven conditions, so that we cannot apply the three additional condi- tions necessary for unit points. CHAP. IV.] SCALAR POINT EQUATIONS. 139 The three points above determined will always be real, i.e. the roots of (342) are always real; for, suppose one to be im- aginary, and call it n -f- n'i, and the corresponding value of p, p +p'i, in which i V 1 ; then or, equating separately to zero real and imaginary parts, p = np n'p', p' = n'p -f- np'. ' P'l^P = np'\P n'p'-=p\p' = n'p* -f- np\p'. which can only be satisfied by u' = 0, so that there can be no imaginary value of n or p. 113. Canonical form of p\p. With the form of < given in (345) we have p\4>p = n l (p\p i y + n. 2 (p\p. 2 y + nz(p\p. A y, . (346) which is the canonical form of the scalar quadratic in p, and we have shown that, whatever may have been the original form of p\p, it can always be reduced to this canonical form by properly choosing the reference points. The first two terms of (346) may be written n s ) or p\q -p\q 2 ) if we put q l =PJ-\/WI -\-p-2\f n-2 and q., = Thus the equation of the locus becomes = ........ (347) In this form the eqiiation shows that the curve is tangent to l^ and \q. 2 at the points where they are cut by \p s . But, by the last article, p^ = 0, and p 3 \q s = 0, so that \q v and q 2 pass through pg, and touch the curve where it is cut byj\pj \p 3 . 140 DIRECTIONAL CALCULUS. [ART. 114. 114. Condition that p\P=P'h-J"h, whence p = K?i ' p\q a + q, ' P\pi,

    p 3 will all be on the line q^o, so that their product will be zero. Conversely, if we have fcPi<&>*&>3 = ......... (348) for any three points whatever, p l9 p 2 , p 3 , then the function < must have the above form, and p\(f>p is factorable. The ex- pression pip-2ps is the discriminant of p\$p, and is the same as pipzPs ' & of eq. (343) . 115. Nature of the locus at oo. To deterinine this we will find the intersection of the line at oo with the locus. In (327) put EI and e 2 for q l and q% thus obtaining V _ eilfog V Now, as Ci and c 2 niay be any points at oo whatever, write = e 1 GO, c 2 = 63 e o '> then e. 2 e so that the quantity under the radical becomes Thus the two vectors Ci!^c 2 + V and e^ca-cj (e^eo+V 3ee l( /) 2 )e, . . (349) are respectively in the direction of the two points at oo of the locus. These vectors are imaginary, parallel, or real, accord- ing as ep = %e ' p\e + ^ - p^ + & 2 ' . p\e 2 , . . (354) and (351) becomes (355) 117. Curve through the reference points. Let A = B C = in (351) ; omit the primes and divide by 2, and the equation becomes Ap\e v - p\e 2 + Bp\e. 2 . p\e + Cp\e p|e, = 0. . (356) 142 DIRECTIONAL CALCULUS. [ART. 118. As this equation is satisfied when p is replaced by e , e l} or 62, it follows that the curve passes through the reference points. It is the most general form of the equation of a conic through these points, because no other term can be formed which will vanish under each of these conditions. It is the same equation as (246) if L^ L 2 , L 3 be taken as sides of the reference tri- angle. The equations of the tangent lines to (356) at the reference points are p\e = 0, etc.; and, by (353), e = Ce l = Ae 2 Hence these tangent lines may be written B C C _ , _ A "" A B (357) 118. Circle through the reference points. drawn through the reference points as in the figure, and also the tangents at these points. Let c^, a 1? 03 be the an- gles of the triangle as shown, and a Oi, Oa be equal respectively to T(e l e 2 ) ) T(e 2 e ), and T(e e } ). Let q be any point of the tangent at e 2 , and let g 2 ~~ } i~q = 0. 120. The equations p\p = C and p\p = represent conies which are concentric, similar, and similarly placed. For, since 3p|e = l, we may write the first equation p\p = C(3pe) 2 without changing its meaning ; but pje = is the equation of CHAP. IV.] SCALAR POINT EQUATIONS. 145 the line at x ; hence the equation in its present form is that of a conic tangent to p\p = at the points where it cuts the line at oo ; that is, the two curves have the same asymptotes, real or imaginary, which proves the proposition. It will also appear on examination that the expression for the center will be the same for the two curves, if we note that c|e is always zero, being the product of a point at oo into the line at oo . Anti-polar of any point. Let e be the point whose anti- polar is to be found, and let e' be so situated that q c = (e+e') ; i.e. the center of the curve is midway between e and e' on the line joining them. Then the polar of e' will be the anti-polar of e ; and since, by the given relation, e' = 2 q c e, we have for the required equation p\(2q c -e) = .......... (359) 122 Reciprocating ellipse. We proceed now to find the equation of the ellipse referred to in Arts. 41-44, with refer- ence to which each reference point is the anti-pole of the opposite reference line. When the reference triangle is equilateral, the reciproca- ting curve is a circle. Now the equation of a circle through the reference points is, in this case, because a,, = % = a* Hence, by Art. 120, the reciprocating circle will be p\e 1 'p\e. 2 -\-p\e 2 -p\e +p\e -p\e l = C(3p\eY . (360) if C be properly determined. We may infer that the equation of the reciprocating ellipse will have the same form when e ei 2 is not an equilateral triangle. When A = B = C = and A' = B' = C' in (351), we have q e = e, so that (359) becomes p\P = K e i + e 2 whence (2e CQ) = 0, or \e (2e e () ) = 0. That is, \e (f>(2e l + 2e 2 e ) = \e (e. 2 + e 0(7e + e -f e l Whence C=%. It is evident from the symmetry of the equations that the equations |e 1 ^>(2e 6^ = and \e 2 <}>(2e e 2 ) = will give the same value of C; hence the equation of the reciprocating ellipse is 6(p\e l 'p\e 2 +p\e. 2 .p\e <) +p\e (> .p\e 1 )-(3p\ey = 0. . (361) 123. Complement of any point. We will give now the proof, referred to in Art. 44, that \p is the anti-polar of p with refer- ence to (361). Let p = Ie + we! -|- we 2 ; then its anti-polar is \(2e - Ie me l ne. 2 ) = |[(f - O e o + (|- O e J + (I - )c 2 ], being derived from (361). We find e 2 = 3(e + z\ e}- NOAV, if the proposi- tion is true, we ought to have (Ie + me, + ne a ) ^[(2 - 3 Oe + (2 - 3 wi) e i + (2 - 3 H)^] = 0. .-. \(le + mei + we 2 )[(2 -3t)(e l + 2 _ g) + (2 - 3m)(e, ei + ne)(3 Ie + 3 ftie^ + 3?ie 2 ) = 0. Hence the proposition is demonstrated. CHAP. IV.] SCALAR POINT EQUATIONS. 147 124. Line equations. If, in eq. (356), we write L for \p, we have AL^ Le, + BLe, Le + CLe - Le l = 0, . . (362) the equation of a conic enveloped by L, and tangent to the sides of the reference triangle, because it is satisfied when L = e^z, or e^o? or e^. Write $L = | (Be 2 + Ce.) Le + \ ( Ce + Ae^ Le l + \(Ae 1 + BeJ-Le S) . (363) and (362) becomes |fL = .............. (364) Comparing with Arts. 116 and 117, it appears that (365) We can thus pass at once from any point equation to its complementary line equation, or the reverse. The function ^ is a linear line function of L, self -con jugate, and therefore possessing all the properties shown to belong to . Differentiating (364), we have dL^L + L\^dL = 2 dLtyL = 0. Hence, by Art. 79 and eq. (364), \\j/L is a point on L, and also on dL, a line through e and the point of contact of L with the curve. Consequently \\j/L must be itself the point of con- tact of L. If L be not tangent to L\\j/L = 0, then \\j/L is the pole of L. 125. Center of L\\j/L = 0, iff being any linear self-conjugate line function. The center of any conic is the pole of the line at oo. Hence mq c \$\e = e. 148 DIRECTIONAL CALCULUS. [A RT . 126. .-. g c = p = pass through e, the center of the reciprocating ellipse. Thus, when a curve passes through the mean point of the reference triangle, its reciprocal curve is a parabola. 126. Determination of the curve L\\j/L = 0. If real tangents can be drawn to the curve p\p = through e, then the recipro- cal curve (364) must have two real points at oo, viz. : the anti- poles of these two tangents, and must therefore be a hyperbola. If wo real tangents can be drawn to p\$p = through e, then (364) has no real points at oo, and is therefore an ellipse. If two coincident tangents can be drawn, the curve is a parabola, as was shown in the last article. Now (330) was shown to be the equation of the tangent lines to p\<^p = through q v ; hence, putting e for q 1} the tangents through the mean point are = = e\e or, writing fap = p 'e\e e-p\e, .... (368) the equation becomes ........... (369) The two lines represented by (369) will be real, coincident, or imaginary, according as they cut the line at co in real coin- cident or imaginary points ; that is, by (350), according as ei f i\t2 is negative, zero, or positive. Hence we have the criterion for eq. (364), + for an ellipse for a parabola for a hyperbola (370) CHAP. IV.] SCALAR POESTT EQUATIONS. Lete 149 127. Pascal's theorem. to find the locus of p when q } , q.^ q s are in one straight line, as shown in the figure. The condition for this 18 WJtfs = 0, and we have q., = Hence e s , be any five fixed points ; ee e 5 p) = 0. (371) This is a scalar equation of the second degree in p. It there- fore represents a conic. It is evidently satisfied when p = e 1? and when p = e 5 . Let p = e 2 ; then q l and q 2 are each on the line e^ and q 5 is the point e 2 ; hence the equation is satisfied. Let p = e 3 ; then g 1 and q 3 each coincides with e s , so that 5^2^3 = 0. Finally, Iet_p=-e 4 ; then q 1 coincides with e^ and q. 2 and q z are both on the line e 4 e 5 , so that q\q& s in this case also. Thus the conic passes through the five fixed points, and the hexagon p, e^ e 5 is inscribed in a conic. We have, there- fore, the following theorem : If a hexagon be inscribed in a conic, the pairs of opposite sides intersect each other in three collinear points. 128. Brianchon's theorem. This theorem is the complemen- tary, or reciprocal, of Pascal's, and will be obtained by writing lines for points in (371). It may be stated as follows : If a hexagon be circumscribed about a conic, the three lines joining the opposite pairs of vertices will pass through a common point. 129. Inversion of <. Let

    ~ l e, and let c be the function conjugate to , so that 150 DIRECTIONAL CALCULUS. [ART. 129. q\(jjp = i)\ c q. Also, let q 1} q 2 , q s be any three non-collinear points. Then (h\P =P\^i = to\ e > q 2 \P In eq. (102) put c c q 2 for p 1} and < c g 3 for p. 2 ; then, noting the equations just given, we have - l e=p= - [jtete Sfi!e + IMg&Sh g 2 je |. (372) When ^> is self-con jugate, of course <^ c = <. If e = (372) becomes As the sum of any two linear functions is itself a linear function, and its conjugate is the sum of the conjugates of the two functions, we can invert such a function by (372). Take for instance (< + n)p, of which the conjugate is (< c + n)p } and we have \( c +n)q 2 ( c +n)q 3 P\q 1 + \(^ e +n^q s (^ c +n)q l -p\q 2 _ ftb^-fo + n x p + Jc 2 n 2 -j- n 3 in which Jc , Jc^ k 2 have the values given in (343) with < c sub- stituted for , and -%P ' l a linear function of p, the coefficient of n in the expansion of the numerator of the second member of the equation. Clearing of fractions and operating by < + n, we have This equation must be true for all values of n, and therefore the coefficients of different powers of n must vanish ; hence ~ 1 p-fciP = and - 1 p = (k 1 Jc 2 + 4> 2 )p ........ (374) CHAP. IV.] SCALAR POINT EQUATIONS. 151 Thus we have another inversion formula. If we substitute the value of x just found in that of (< + w)" 1 as written above, we have a formula sometimes useful, viz. : Finally, operating by <, (374) may be written (< 3 - ^ + krf - Jc )p = ....... (376) 130. EXERCISES. (1) Invert $p as given in eq. (352) with A = B - C= 0. Also with A' = B' = C' = 0. (2) Invert <]>p= nip + e x -p\e 3 Ans. ~ l p = [mej ^[ej + (e z - ej -p\e 2 + em p|e ]. 77^ (3) Show that if the sides of a triangle pass through three fixed points, and two of the vertices move on fixed right lines, then the third vertex describes a conic. Ans. Equation of conic i (4) Show that if the vertices of a triangle move on three fixed right lines, and two of the sides pass through fixed points, then the third side envelops a conic. Ans. Line equation of conic is LL^p^p^L = 0. Pascal's and Brianchon's theorems can be derived from (3) and (4) respectively. (5) When the three points of Ex. (3) are collinear, show that the locus reduces to two straight lines. (6) When the three lines of Ex. (4) have a common point, show that the envelope reduces to two points and the straight line joining them. (7) Write the equation of a conic passing through four points and tangent at one of them to a given right line. Ans. ei - e 3 e 4 e^g e 4 c 4 e^ z e 4 p = 0. 152 DIRECTIONAL CALCULUS. [ART. 130. (8) Write the equation of a conic through three points and tangent at two of them to given lines. Ans. (pe l e 2 e s ) (e^ e 3 e 3 ) (e^ - e 4 p) = 0. (9) Write the line equation of a conic tangent to four right lines, the point of contact being given on one of them. Ans. Lines are L u L 2 , L 3 , e 4 e 4 , and point e 4 , and equation is (L 5 e 4 c 4 )] {L,L, . ej [ L 2 L 3 . ( A - L) ] = 0. (10) Write the equation of a conic tangent to three given lines, the points of contact being given on two of them. Ans. If lines are e^ L 2 , e 3 c 3 , and points e l and e 3 , then the equation is [ (Le^Lj^]e&[(e& L 3 ) (e 3 e 3 L}~] 0. (11) If L, p, <, and if/ are related as in Art. 124, show that L\\j/L = and p\$~ l p = are the line and point equations of the same curve, viz. : the anti-polar reciprocal of p\p = 0. Also that L\I/~ 1 L^=0 and p\p = Q are the line and point equations respectively of the anti-polar reciprocal of L\\frL = 0. (12) Show, by (328) and (373), that the condition that A shall touch the conic p\p = 0, is L^' 1 L^ = 0. (13) Interpret the complementary condition puf/'^pi = 0. (14) Find the locus of p under the following conditions: e o> e D e o> \ are fixed points ; c and ej are given vectors ; p' = e ' + 0*0, p" = ti + 0*1, and e p'p e^'p = 0. Ans. e Q e 'p (15) Write in the result of the last exercise e " e ' for CQ, and Ci'ej 1 for e^ take the complementary equation and inter- pret it. (16) The sides of a triangle cut the corresponding sides of its polar triangle with reference to any conic in three collinear points. Reciprocally, the lines joining the corresponding ver- tices have a common point. CHAP. IV.] SCALAR POINT EQUATIONS. 153 (17) Given four points ; through them, two by two, draw six lines, cutting each other in three additional points, say q lf q 2 , q s ; then will any one of the g's be the pole of the line through the other two, with reference to any conic through the four given points. (18) Derive the reciprocal proposition. (19) Find the conditions that a triangle inscribed in a conic shall have maximum area. Let p, p', p" be the vertices of the triangle, subject to the conditions The area is u=pp'p", and for a maximum or minimum du=0. .: dpp'p" + dp'p"p -f dp"pp' = 0. But the p's are independent of each other, so that we have dpp'p" = dp'p"p = dp"pp' = 0, and also dp\P = dp'\p' = dp"\^p" = 0. Hence p'p" is || to |!(e + e 1 + e 2 + e 3 ) = 4p;e = l ..... (377) The equations of curves and surfaces in solid space may appear under the following forms : Expressed in terms of points Non-scalar equations. Scalar equations. Expressed in terms of planes Expressed in terms of vectors Expressed in terms of points Expressed in terms of planes Expressed in terms of vectors 132. The non-scalar equation p = 2o(ze) = CQ + 2 (378) 164 DIRECTIONAL CALCULUS. [ART. 133. cu ca, e 3 having the values given in the last article, and x being eliminated from the third member of the equation by the rela- tion x -f- x l + x 2 + Xg = 1, may be called the equation of solid space, since, by giving proper values to XQ, x^ etc., p may become any point whatever in space. If a relation be given between the scalar quantities, such as fi(x<>, a* ** a*) = 0, or f^x,, x,, z 3 ) = 0, then p will lose one degree of freedom of motion, and will lie on some surface. If another relation be given, such as ft(*a> *i, * ZB) = 0, or / 2 (Zi, x,, x s ) = 0, then p will be compelled to move on a second surface simul- taneously with the first, and hence along the curve of inter- section of the two surfaces. It follows that the non-scalar equation of a surface has two independent scalar variables, while that of a curve has only one. Eq. (378) may be written in the form p = e 9 'p\e t + e l 'p\e l + e,-p\ei + e s -p\e m . . (379) from which it appears at once that the scalar coefficients are proportional to volumes of the tetraedra formed by joining p with the reference points. Since p is a unit point, the truth of (377) appears also from (379). It is easily seen likewise from (379) that when p passes through any face of the refer- ence tetraddron, the coefficient of the opposite point changes sign. Thus when the coefficients are all positive, p is inside the tetraedron ; when one is negative, it has passed through one face ; when two are negative, it has passed through two faces, i.e. through one edge; and when three are negative, it has passed through three faces, i.e. through one vertex. 133. If P \p, then the complementary equation to (378) or (379) is P=$l(x> e)=2o(|e-eP), ...... (380) and P may be any plane whatever in space. But if a relation exists, as in the last article, between the scalar variables, then CHAP. V.] SOLID GEOMETRY. 165 P moves according to some definite law, and envelops a sur- face. If a second relation exists between them, then P touches two surfaces simultaneously, or rolls on them, and therefore envelops a developable surface, which is reciprocal to a curve. 134. Writing in (378) p e = p, we have />-=3!(0, ........... (381) a vector equation which, regarding p as always drawn outwards from a fixed origin, represents a surface or a curve under the same conditions as previously given. 135. Equations of planes, lines, and points. If, together with eq. (378), we have the linear relation S2(maj) = 0, .......... (382) then (378) represents a plane; for, since x =p\e , etc., we have e) = 0, ...... (383) which is the condition that p shall be always on the plane | (m e + mfr + m^e 2 + m s e 3 ) . If we have also the relation Sj(na;) = 0, ............ (384) p must also lie on the plane whose equation is ^(ne) = 0, .......... (385) and hence must lie on their common line. The equation of a plane through any three points p^ p 2 , p s may be written P'*p l + x(p-pi) + y(ps-pi), . (386) of which the scalar form is Q, .......... (387) 166 DIRECTIONAL CALCULUS. [ART. 136. which may be derived from the non-scalar form by multiplying it by pipzPs- The equation of a plane through p l and || to P l is (p-p 1 )P 1 = 0; (388) for it is satisfied when p=p\, and is the condition that the vector p pi shall be || to Pj. The complementary equations to (383), (385), and (387), viz. : P2,(nie) = Q P20ie) = PPP>P =0 are the conditions that the variable plane P shall always pass through the fixed points 2(wie), 2(ne), and P^PoP^ and hence are the plane equations of these points. Eqs. (233), (234), and (235) are the equations of lines in solid space as well as in plane space, though in the former they are all non-scalar. 136. Vector equations of planes and lines. The equation %l(Ax) = <7, (389) taken in connection with (381), represents a plane, if p be always drawn outward from a fixed point ; for, eliminating x 3 between the two equations, we have + A* I -""I i> *i e i - -r- (390) or the non-scalar and scalar forms of the equation of a plane through the end of Ce s -*-A 3 and parallel to the vectors A& A& and A 3 f 2 A.f%. If a third equation, %l(Bx) = C', (301) be given, we have, eliminating o^ and x.,, Cj , 2 , \, A 2 , x 3 A 3 C xB-C' = 0, or (392) an equation which evidently represents a right line. CHAP. V.] SOLID GEOMETRY. 167 Eqs. (390) and (392) are of the respective forms p = e + a*' + ye" I (p _ e ) V' = L ........ (393) which are those practically used. The equation of a plane through the ends of ej, e 2 , c 3 drawn out from the origin may be written at once in either of the forms or ( p _ l ) (2 _ i ) (3 _ ei) = the last being equivalent to P( 2 e 3 + c 3i + eie2) = ei23> ...... (395) as will be found on multiplying out. 137. EXERCISES. (1) What is the meaning of the two equations Pp^ = and Pp 2 = taken simultaneously ? (2) Interpret the following equations written in comple- mentary pairs: f PPl^2 = ) | PPiP& = ) f > 2 AA = ) ^O)' \PP 1 P 2 q l q 2 e = 0} ) I PL, L lPl = j ' P 3 P 4 P 5 = ) ( p^P! + Vi) = ) = o r ( Pfc 1 i> 1 + te -or j P(> 2 -Pi) PiP* = 1 ip 2 p 3 = C and ppip 2 pi = C' is (Pa~ Pi) (PiPzPsPi 'Pi +C(Pil ) i) C'(P 3 pi))- (6) Show that the common point of the three planes, *n T> n C 11 f>n n n C 111 i 2 2 s2 } 2 s/ \L e + C'(e p\) + C"(e p 2 ) + C'"(ep 5 }. (7) Show that Pip 2 p s = is the condition that the three planes p\pi = p\p 2 =p\p 3 = shall have a common line. (8) Find the condition that the plane p\p 4 = 0, together with the three planes of the last exercise, shall have a common point. (9) Show that if the equations of three planes, on being multiplied by any constants and added, vanish identically, i.e. for all values of p, then the three planes pass through a com- mon line. Also that if the same holds for the equations of four planes, then the planes have a common point. (10) Show that when Pip 2 p s p t = C l C 2 + C s C i} the four planes pp\p 2 p 3 = C^ ppzPsP* = C 1} pp^P\P\ @2> PP*P\Pz @3 have a common point. (11) Interpret the equations /a|e = (7, pee' =(7, (p )|e = 0, (12) Find the vector perpendiculars from the origin on the planes of Ex. (11). Ans - "T> 7~rS' c > Cl T/ ' T*!^ 3 * (ee')= ((2 ejcs)- (13) Find the conditions of perpendicularity and parallelism of the two planes p|e = C and p|c' = C'. CHAP. V.] SOLID GEOMETRY. 169 (14) Find the vector perpendicular from the end of the vector 8 upon the line (p e)c' = 0. Ans. e'(e 8) ['. (15) Find the conditions that the three lines (p e)e' = 0, P l P. 2 p = 0, Pipzp = shall lie respectively in the three planes Ans. e'|e" = and e\t" = C, (P* Pi) \Pz and p 2 \p s = C. (16) Show that the shortest distance between the two right lines (p-e,)e 2 = and (p- ei ')e 2 ' = is ( g i~ c i')W. (Use Art. 46.) T ** 2 (17) Show that the equation of the common line of the two planes (p e)e'c" = and (p eiXV = is (18) Show that the equation of a plane through the line (p c i) e a = parallel to the line (p fi)e 2 l may be writ- ten (p ej)e2 W = 0, and hence, by Ex. (17), that the line ,f f > -I- V\ f c ' "(W) 2 2 H cuts these two lines at right angles. (19) What are the conditions that the pairs of right lines JfcfcC*-*,) =0| and J-pi) = oj ( ( P _ ei ') e2 ' = o) -4- Pi (ft - 9i)Pi'(Qa ~ ft') = and ( ei - e 1 ')e 2 e 2 r = 0. (20) Show that the common point of the three planes p\^i = Ci, p|e 2 = (7 2 , p|c 3 = C r s is at the end of the vector 170 DIRECTIONAL CALCULUS. [ART. 138. (21) Show that the three planes of the last exercise will have a common line if we have = and Cifte + (7 2 3 c i + C&& = 0. (22) Show that, if CjeoCsC^ e-fstiCi + e&\C z e^i^A = 0, the plane p]e 4 = C 4 , and the three planes of Ex. 20 have a com- mon point. (23) Interpret the equations /( P iO=0, /(7=0, /(^) = 0; if /is the symbol of a scalar function. (24) Three planes pass through the three lines of intersec- tion respectively of the three planes (p- 1 )[e 1 = 0, 0>-e 2 )|e 2 = 0, and (p-c 3 )| 3 = 0, each being perpendicular to the opposite plane (i.e. the plane through the common line of the first and second planes is per- pendicular to the third, etc.) ; find the conditions that the three first mentioned planes may intersect in a common line through the origin. (25) Find the shortest distance between the diagonal of a cube and an edge that it does not meet. (26) Find the equation of a line through p l cutting L^ and L^ Ans. ^iL- = 0. (27) Derive (395) from (387) by transformation, as in Art. 75. 138. Tlie sphere. The equation p = a[ (ti cos 6' -f i 2 sin 6') sin 6 + 1 3 cos 0] } > . (o9o) = tt(rsin0 + i 3 cos 6), say, ) so that ......... (399) CHAP. V.] SOLID GEOMETKY. 171 represents a sphere. For, taking the co-square, we have p? = a 2 (r ? sin 2 -f- 1 3 ? cos 2 6) = a 2 , since r\i^ = and tg 2 - = T 2 = 1. Hence 7p = a = constant, which is a property of the sphere with center at the origin. If the center be at the end of e, we have for the scalar equa- tion T(p c) = a ; whence p l _ 2 p ' = a- - c 2 - or p|(2e - p) = e 2 - - a 2 . . (400) This equation is identical with (269), so that it represents a circle or sphere according as it is interpreted in plane or solid space. The properties of the plane-radical can be proved pre- cisely as, in Art. 82, those of the axis-radical were demon- strated. 139. EXERCISES. (1) If a and /3 are any two non-parallel vectors, show that the equations P 2 = ft 1 ( p ;a+C 1 ), f^ = k,(a(3p + <7 2 ), pl = k 3 (a(3\a.p+C 3 ) represent spheres ; find their centers and radii, and their rela- tive positions. If Ci = C. 2 = C 3 = 0, show that they cut each other orthogonally. (2) If a, /?, y, 8 be any four vectors drawn outwards from a point, and the relation a/3y - (3yS - a? + ySa & - Sap y? = exists between them, show that the extremities of the four vectors and their common point all lie on a sphere. (3) Show that the equations of the tangent plane and nor- mal line to (400) are respectively e) a =H, ....... (401) ( c i|oa = 0> etc., pjaj = x, p|o2 = y, and p|a 3 = whence (pi^) 2 + m(p\a 2 ) 2 = a p \a 3 = P \ P ; .... (404) in which , (412) p = T' cosec + ci s cot } in which T = a^ cos 0' + 6i 2 sin & ,}, -.. (413) T' = at! sec 6' -f 6i 2 tan 6 are called respectively the ellipsoid, the liyperboloid of one sheet, and the hyperboloid of tivo sheets, for reasons which will pres- ently appear. CHAP. V.] SOLID GEOMETRY. 175 By comparison with eqs. (276) and (277) it will be seen that eqs. (413) represent respectively an ellipse whose semi-axes are a and &, and a hyperbola with the same semi-axes, each lying in the plane i^ the vector radius of one curve being T, and that of the other T'. In the first of eqs. (412) give to T some particular value consistent with (413) ; then the equation represents an ellipse whose semi-axes are TV and c ; hence the first of (412) represents a surface generated by an ellipse re- volving about i 3 as an axis, c being the semi-axis along i 3 , while the other semi-axis is the radius-vector of the ellipse in t^ whose semi-axes are a and b. Similarly, the second of (412) represents, for any given value of T, a hyperbola whose semi-axes are TV and c, and, when T varies subject to ^413), it represents the surface generated by this hyperbola revolving about tg, having c for its semi-axis along i 3 , while the other is the radius vector of the ellipse in iii 2 - Finally, the third of (412), for any given value of T', repre- sents a hyperbola with semi-axes TV' and c, T' being, as we saw above, the radius-vector of a hyperbola in i^ Thus the sur- face is generated in this case by a hyperbola revolving about i 3 , having its c-axis constant, while its other semi-axis is the radius-vector of a hyperbola in L^ whose semi-axes are a and b. The methods of generation of these three surfaces show that the first is a limited surface, having no real points at oo, and hence that no plane can cut from it a hyperbola or parabola ; hence the name ellipsoid : that the second is generated by a hyperbola in such a way as to form one continuous surface, so that any two points whatever lying on it can be joined by a line also lying wholly on it-, hence the name hyperboloid of one sheet : finally, that the third is generated by a hyperbola in such a way as to form two distinct portions, or sheets, so that points on these respective portions cannot be joined by a line lying wholly on the surface ; whence the name as given above. 176 DIRECTIONAL CALCULUS. [ART. 144. 144. Eliminating and 6' from eqs. (412), we obtain =)- m--(TJ=U, .... (4: 'P\^ the scalar equations of the same surfaces. As an exercise let the student determine the nature of the sections of the surfaces of (414) by the three planes pl^ = C 19 p\L 2 = C 2 , p\is = C 3 when d > a, C 2 >b, C s > c, C = a, C 2 = b, C 3 = c, C, < a, C 2 < b, C ?J < c. 145. Write then the equation , = i HJ3. .... (415 ) or b 2 c- l- ........... (416) is equivalent to any one of the equations (414), if we select the signs properly in (415). The function < as given in (415) is evidently self-conjugate, and possesses all the properties proved in Arts. 86 and 89. Eq. (416) being of precisely the same form as (282), any operations performed on (282) which did not depend on the form of < will give identical results when performed on (416), the interpretation of the results being, of course, different. The discussions which follow hold for any linear, vector, self- conjugate form of <, except when is specially restricted to the form (415). 146. Tangent plane and normal. By differentiation we have, precisely as in Art. 87, that (f>p is parallel to the normal at the end of p, and hence that the equations of the tangent plane and normal line at the end of p are respectively and ( t r p)< P p = 0, ........ (418) CHAP. V.] SOLID GEOMETRY. 177 which are identical in form with (284) and (285). Also, as in the same article the perpendicular from the origin on the tangent plane is 7~ U+p ....... ; - (419) lp 147. Diametral plane. Repeating exactly the operations of Art. 88, we obtain the same equation o-|4>c = 0, .......... (420) which now represents a plane, the locus of the middle points of a system of chords parallel to c. This plane is perpendicular to p 1 ; that is, (421) then a, ft, y are a set of conjugate semi-diameters. From these relations we have a = m\(f>fty ; multiply into |afty. Substitute value of m, and we have a -y= \fty 1 and, similarly, ft aft(}>y = | ...... (422) y a.afty=\fty, ft afty = \ya, y afty = \aft. (423) 178 DIRECTIONAL CALCULUS. [ART. 148. 148. Interrelation of the equation o-|^> = l. In the first place it evidently represents some plane parallel to that of (420), i.e. conjugate in direction to e. If we have e| = 1, with the condition e|

    = l ........... (423a) represents a plane containing the points of contact of all tan- gent planes to the surface which pass through the end of e, or, in other words, the plane of the curve of contact of the cir- cumscribed cone whose vertex is at the end of e. Of course the end of e may be so situated that this cone is imaginary. The plane f 2 = 1 respectively ; if Pi be on the polar plane of p 2 , we must have the equation satis- CHAP. V.] SOLID GEOMETRY. 179 fied when o- = c 1 ; i.e. e 1 |P = p'<- P\r? = ( or T-p = l; ..... ..... (424) so that, when p is a vector of the surface, <-p is always a unit vector. Also, if a, (3, y are vector, conjugate semi-diameters, a|?lp = ^/3\^a = 0, and, similarly, < 2 /3j<' v y = '-y|< 2 a = 0, so that < 5 a, -/?, -y form & unit, normal system of vectors. 150. The volume of the parallelopiped formed by tangent planes at the ends of a set of conjugate diameters of an ellipsoid is constant. Taking a, /?, y as in the last article, the required volume is 8a/3y. Now with <, as in eq. (415) with the upper signs, we have 2 y = */3 + etc. i >-y + etc. ? (425) = 8a6c = 8 a&c, by the last article and equation (195). Take the co-square of each of equations (425) and add them, and we have a ? + /S 2 - + y ? = a 2 + Zr' + c 2 (426) 180 DIRECTIONAL CALCULUS. [ART. 151. 151. The equation (Pypr(yapr-(a(3 P y = (a{lyy .... (427) represents a central quadric referred to the conjugate semi- diameters a, ft, y. The origin is at the center because the equation is unchanged by putting p for + p. If a = ai 1} ft = 5i2, y = cig, the equation becomes identical with (414) Eq. (427) is satisfied when p=a, or p=/?V 1, or p=yVl, so that, when the signs are all positive, the ends of a, /?, y are real points on the surface; when one sign is negative, one point becomes imaginary ; and when two signs are negative, two points become imaginary. If we write p = yap + la/3 aft,), (428) eq. (427) becomes pj = 1, and we have at once the conditions for conjugate directions. 152. Rectilinear generators. Eq. (414) may be written in the form = 1+- a 1- or a?p\a - p\a' = (a + pi^ (a , . (429) which shows that the surface passes through the four lines of intersection of the pair of planes p a = and pa' = with the pair of planes p\ii = a and p]^ = a, Now if we take the upper signs throughout, a and a' are imaginary ; hence for the ellipsoid these lines are imaginary. Again, taking the lower signs throughout, a and a' are imaginary, and the lines are therefore imaginary for the hyperboloid of two sheets. If, however, we take the upper sign for the second term of (415) CHAP. V.] SOLID GEOMETRY; 181 and the lower sign for the third term, or vice versd, then a and a' are real, and the four lines of intersection of the two pairs of planes are real lines on the surface, which is now a hyper- boloid of one sheet. In this case we have a = b~\ + c" J i 3 and a' = &~ ] i 2 c" 1 ^. Either of the two pairs of planes, ,!a+- tl ) = / p\(aa m 1 ) = na\ (pK ^ -f a - and - i , (429a) (aa-- tl )=- J taken simultaneously satisfies eq. (429) . Hence the two planes of each pair intersect in a line lying wholly on the surface. By varying m and n we thus obtain two systems of rectilinear generators of the hyperboloid of one sheet. 153. EXERCISES. (1) Show, by Ex. 22 or 9 of Art. 137, that no two generators belonging to the same system intersect, while every generator of one system cuts every generator of the other. (2) Show that the vector to the common point of a genera- tor of the system n and one of the system m is p = (m -j- n)~ l [a,i l (m n) + bi(mn + 1) + ct 3 (mn 1)]. (3) Find the condition between m and n in order that a generator of one system may be perpendicular to one of the other. , , ,,/! VI > "f 1 i V 1 j. "\ n Ans. 4 cr b~l -- m}( -- n cr I -- h m [ |- ?i = 0. \m ' J\n J \m J\n j (4) Show that the projections of the generators on the re- ference planes are tangent to the principal sections of the surface ; that is, that the projecting planes of the generators touch the surface at points lying in the reference planes. (5) Show that if a, (3, y be substituted for ai 1? 612, ci 3 in (412) and (413), those equations will be equivalent to (427). 182 DIRECTIONAL CALCULUS. [ART. 150. (6) Find the locus of the intersection of tangent planes at the ends of conjugate senii-diameters a, (3, y. Let o- be the vector to the point of intersection ; then cr = a + ft + y and p is o-\p. (8) Show that, when Tp = k = const., the locus of the end of p is p\

    p, p\(f>p = 1, and p = 1 ; therefore ~ 1 (o- e) = xr, whence (o-- e)^ 1 ^- e)=[o-|(o- -e)] 2 , . . (430) a surface of the fourth order. If we change the origin to the end of c, by writing p for o- e, we have (431) (10) Show that the vectors joining any point on the surface with the extremities of a diameter are conjugate in direction. (11) Find the value of C so that p e = C may represent a plane tangent to p\(f>p = 1. Ans. C = Ve}*/)"^. Compare with (417), and note that the equation is independent of Tt. CHAP. V.] SOLID GEOMETRY, 183 (12) Taking tf> as in (415), and cj, c 2 , c 3 , as unit normal vec- tors, show that 3 = 7- C&2 02 ^2 (13) Find by Exs. (11) and (12) the locus of the common point of three perpendicular tangent planes to a central quadric. Ans. (r 2 = a 2 & 2 c 2 . 154. Condition that p\p shall be factorable. Proceeding pre- cisely as in Art. 114, in fact simply putting vectors for points in that article, we have for the required condition 4>A<^v = 0, (432) A, p., v being any three vectors whatever. The function p <(>p kp- is always factorable ; for, writing it in the form p\( k)p, and putting tf> k for < in "(432), we have (-k)\(-k)n(-k) v = Q, . . . (433) a cubic in k which must have at least one real root : hence a value of k can always be found which will make /[(< k}p the product of two linear factors. 155. The function in general. Any linear vector function may be written in the form ^ = 1 . 1 V + C 2 . e2 V + 3 . 3 V (434) This may be shown precisely as was done in Art. 97 for two- dimensional space. The function c p = e/ cjjp + c 2 ' falp + 3 f 3 |p . . . (435) is conjugate to <; i.e. a- p = p\ c a; and (< + < ( ,)/ ! > is always self-con jugate, as was shown in Art. 97. Again, p\p = p\ c p or P \( - < c )p = 0. .. (< < c )p is perpendicular to p, and we may write 184 DIRECTIONAL CALCULUS. [ART. 156. e being some real vector when < is not self-conjugate. Then we have fr = |(4> + + + g) ~ l p | 1 (442) The coefficients m , m^ m 2 are invariants; i.e. their value is the same whatever X, //,, v may be, which may be shown, as in Art. 98. 156. The general scalar equation of the second degree in terms of vectors. This is identical in form with eq. (312) for plane space, viz. : 2y\p = C- ) ........ (443) for all second-degree terms may be included in p\p, and all first-degree terms in 2y\p. We will first show that the surface represented by (443) has, in general, cyclic sections; i.e. that certain planes will cut circles from the surface. Add and subtract kp- to and from eq. (443), which thus becomes By Art. 154, k can be so determined that p|(< k)p shall be CHAP. V.] SOLID GEOMETRY. 185 factorable ; i.e. it may be written p\a p\a', k being one of the roots of (443). Hence (443) may be written k{ + 2y\p-C + p\a.p\a' = 0,. . . . (444) which represents a surface passing through the curves of in- tersection of the sphere kpr + 2 y\p C = with the two planes p\a = and p\a' = 0. As these curves are necessarily circles, these two planes cut circles from the surface. 157. Let us apply the results of the last article to eq. (414). Eq. (433) becomes in this case, if we put A = ii, /x = ig, (a- - -k)( b~ 2 -k)(c~ 2 -k) = 0, which gives the three values of k. Using the value k = b~ 2 for the ellipsoid first, we have b 2 b 2 or. writing -- 1 = d 2 and 1 -- - = e 3 2 , and reducing, c 2 a 2 or again p\ (e^ + e^O P \ (e^ - e^) = 6 2 p ? , . . . (445) an equation in the form of (444). The two planes = and el - = cut cyclic sections from the ellipsoid ; they pass through the b axis of the ellipsoid, and are inclined to the a axis at the angles tan" 1 i? and tan" 1 ) ! ); i.e. i V V ab*-c> and The values of d and e s are real only when 6 lies between a and c in value. For the hyperboloid of two sheets b 2 and c 2 are negative, and therefore ei and e 2 are real when b lies between a and c in numer- 186 DIRECTIONAL CALCULUS. [ART. 158. ical value, so that the cyclic planes of (445) pass through the greater axis which does not pierce the surface. For the hyperboloid of one sheet let a 2 be negative and 6 2 and c 2 positive ; then e 3 is real, and Ci is real when b is greater than c, so that the cyclic planes of (445) pass through the greater axis which pierces the surface. 158. We will next apply the results of Art. 156 to eq. (404), substituting -^ and i 2 for a x and og. We have then from (433) The root k = shows, as is evidently true, that p p is, in this case, factorable without any addition of a multiple of p-. This leads to the rectilinear generators of the hyperbolic paraboloid as in Art. 141, which are sections of the surface by infinite spheres. Taking next the root k = m, we have, after reduction, the equation l m + t 3 Vm) p\ (ii Vl m 1 3 Vm) j, . (44G) which gives real cyclic sections when m is positive and less than unity. Similarly, the root k = 1 gives p\ (i 2 Vm 1 + 1 3 ) p\ (ijVm 1 t 3 ) = ap\ia fi, . (447) which gives real cyclic sections when m is positive and greater than unity. Eqs. (446) and (447) both represent elliptic paraboloids, so that the only cyclic sections of the hyperbolic paraboloid are the generators as mentioned above. 159. EXERCISES. (1) Show that the two planes p\ (ei8 + e 3 ii) = C and p \ (e^ e 3 = C cut the surface p\p = 1 in circles that lie also on the sphere CHAP. V..] SOLID GEOMETRY. 187 (2) Show that, when C = Va 2 c 2 , the planes of the last exercise touch the surface in the points p = (e^ig e 3 a 2 ii) -s- ( Va 2 c 2 ). These points are called the umbilici. (3) Show that the locus of the common point of tangent planes at the ends of three perpendicular radii vectores is * etc., OTJ^P! = 1, etc., and Pl \p 2 = p 2 \p 3 = p 3 \pi = 0. Hence, \pip* = P ' p2 Ps, etc. J-Ps Now, by (ITS), putting P\Pl + \P2p3 + \p3Pl), or, by (439), = (pip 2 ps) ~ 1 ~ 1 ( pipz + \p2ps + Pspi) . Ps etc.) / / / \ 9 1/9 -*- i J- i -1- -*- i -L -* .-. (Upx etc. (4) Find the equation of a cone with vertex at the end of e circumscribed about the central quadric. -4ns. With the origin at the vertex, the equation is (5) Show that the vector to the pole of the plane or a = C is <& V-r- C. 188 DIRECTIONAL CALCULtJS. [ART. 160. (6) A plane is tangent to a central quadric ; find the locus of its pole with reference to another quadric concentric with the first. Am. If the equations of the quadrics are p] 2 p = 1 and plfap 1 respectively, the locus is p\-'-A-)-V = l .-'.-' ..... (448) have the same foci, being, as in (415), with the upper signs. These surfaces are said to be confocal. (9) Show that through any point in space three confocal surfaces pass, corresponding to three values of Jc, and that these surfaces cut each other at right angles. See Ex. 2, Art. 93. (10) Show that the equation of the tangent plane to the locus of (443) is a\p +y|(o- + p) = C. If p does not satisfy (443), the equation is that of the polar plane. (11) In Ex. (6) substitute for the second quadric, p\ l p+2yp = C, not concentric with the first, and show that the locus is then 160. Center of the general quadric surface. When the origin is at the center of a surface, terms of odd degrees must not appear in the equations, for it must be unchanged when p is put for -f p. If then we change the origin in eq. (443) by putting p + 8 for p, and cause the first degree terms to vanish, the origin will be at the center. Eq. (443) becomes thus or CHAP. V.] SOLID GEOMETRY. 189 If the first-degree terms are to vanish for all values of p, we must have 8 + y = 0, or vX My ~{~ *^Xu. v y s= ~y = JL\JL .JL ' -, (449) ra say, so that m = (X^v) , >.... (4oO) and \l/y = (\fj-v) ~ l (\iJiv - X\y + etc.) ) The quantity m is an invariant, as we saw in Art. 155. The vector \f/y is also an invariant; that is, its length and direction are independent of the vectors X, M, v, as may be shown by substituting IX + m/* + nv for any one of the three. It also appears from the fact that, as (443) represents a defi- nite surface, when < and y are given, its center must be a definite fixed point, and hence 8, the vector to this point, must have a definite fixed value, so that, m being invariant, \f/y must be so likewise. If in (449) we have m = and TtyyO, ........ (451) the center is at oo in the direction of \f/y, and the surface is said to be non-central. But, by Art. 154, m = is the condition that p\p shall be the product of two factors of the first degree ; i.e. that it shall have the form p\^ p\(3 2 - Also, since the cen- ter is at oo, take the origin on the surface by making the con- stant term disappear ; i.e. (452) an equation that gives two definite values for T8 when the direction of 8 is assumed. Eq. (443) thus becomes 0, ...... (453) which is identical in form with (410), and must therefore rep- resent an elliptic or hyperbolic paraboloid. 190 DIRECTIONAL CALCtTLTTS. [ART. 101. If we have 7n = = 2tyy, ...... .... (454) the value of 8 is indeterminate, and S-2v!S=L . ._ . . (4oo) = C", say The sign of C' will be always taken SLS positive. If we have that is, m C + yj^y = 0, .......... (456) then (455) becomes 0, ............ (457) which, being independent of Tp, must represent a cone, real or imaginary, which, when 8 is indeterminate, breaks up into two real or imaginary planes. 161. Maximum and minimum values of Tp. Tp will be a maximum or minimum when dTp = ; i.e. when p\dp = 0. Eq. (455) gives dp\p = Q; hence, at the maximum or mini- mum points, p and p are both _L to all values of dp, which requires them to be parallel to each other ; i.e. we must have p(j)p = 0, an equation whose solution will give the required values of p. This equation is equivalent to ....... (458) CHAP. V.] SOLID GEOMETRY. 191 g being a scalar constant to be determined. Multiply the com- plement of (458) by p, and we have p\p = C' = gp*-, ....... ... (459) which gives the relation between ( 2p) max . and g. By reference to Art. Ill it will be seen that the equation to be solved here is identical in form with the one there treated. Hence it will only be necessary to put vectors, say p, A, /A, v, in place of the points p, q l} q 2 , q 3) in Arts. 1-11 and 112, in order to obtain results for a vector system in solid space cor- responding to those for a point system in plane space. We will also substitute g for n, and m for Jc. Thus we have, from (341), p\(-g)\ = p\(-g)n = p\(-g) v = Q; . (460) from (342), f or g 3 mog 2 + m>\g m = } in which m has the value given in (450), The ra's are, of course, invariants like the k's of Art. 111. Eq. (461) is called the discriminating cubic, and plays an im- portant part in the discussion of eq. (455). Eqs. (459), (460), and (461) give the complete solution of the problem in maxima and minima; for (461) gives three values of g, say g l} g 2 , g s , which, substituted in (459), give the lengths of the maximum or minimum radii vectores, and substituted in (460) give the directions of the same. If we assume /3 = <733> ...... (464) 192 DIRECTIONAL CALCULUS. [ART. 162. in which a l5 0.% a 3 are a unit, normal system of vectors, so that, whatever may be the original form of , it may always be transformed into 4>P gi a l ' P\ a l + 9 f 2 a 2 * Pi a 2 -\-9sfl-S ' P a 3> .... (4G5) in which a = a = a/ = 1, and a^a, = aslag = a^^ = 0, as well as a]| y * Coincident planes. 01 /o + * Hyperbolic cylinder. 01 10 + - + Two intersecting planes. 01 lo - + Elliptic paraboloid. + CO CO + + sP Parabolic cylinder. 00 00 + Hyperbolic paraboloid. - 00 CO - + 165. EXERCISES. (1) Find what surface is represented by the equation (pe 1 ) 2 + 2p|e 1 zero, and also when it vanishes. 2y|p = C, when is not (2) Discuss the equation p- n 2 (p|ti) 2 +2|o|y=(7 when ?i>l, when n = 1, and when n < 1. 196 DIRECTIONAL CALCULUS. [ART. 165. (3) Discuss the equation p!i p|0i + pK p|02 + p| 3 f>|03 + 2 />!y = C' in the following cases : (a) a 1 = 5i 1 + 2i 2 -j-4t 3 , 02 = 21! 2i2 3i 3 , a 3 = 4ij 3i 2 1 3 , 01 = 'l, 02 = 03 = '35 7 = 0, C = 1. (5) 3 = a t , 0! = a 2 , 2 = a 3 , C= 2a 2 , y = 0. (c) The same as (6) except y= c^ -f- 3o2 lo^ (d) The same case as (6) except add 2/- to the first member. (e) ai = 0i = ai 2 + &ij, a 2 = 02 = 5i 3 + C'2> s = 03 = Ct i y = 0, (7=1. (4) Discuss the following equations : (a) (e ]/ ,)? + (e 2P )? + (e 3 p)?=a 2 . (6) Cip 363 tgp = C 2 . (c) (ejp C 2 t 3 )e 3 (2p Cgq) = C 2 . (e?) epfo =;C, being as in (415). In the last case we may write = (7. (5) Show that, if p\ip = 1 and p\2p = 1 have parallel axes, then p\(i + & 2 )p = 1 has its axes parallel to theirs. (6) Find the condition that the surface of the second order passing through the common line of two quadrics shall be developable, and thus show that in general four different cones pass through this common line. (7) Find the axes of a central plane section of a central quadric ; also the area of the curve of section. The equation of the surface is pj

    p = 1 = kp-. Also, p = k' (< + k) ~ l , whence e\p = = e| ( + &) -1 - By (441) this is equivalent to m t\~ l f. + k* (m 2 )f. + k 2 <2 Q. Area = -n-a = ^ = v if Te = 1. (8) Perpendiculars are drawn from p on the four faces of a tetraedron, the feet of the perpendiculars being coplanar ; find the locus of p. (9) Find the locus of a point, the sum of the squares of whose distances from n fixed points is constant. (10) Find the locus of a point, the sum of the squares of whose distances from two fixed right lines is constant. (11) Find the locus of a point, the sum of the squares of whose distances from three fixed right lines is constant. Determine the nature of the locus in each case. (12) Find the condition that the plane in the tangent plane at p. Then, if P = \p, P will envelop some surface, and we have dP [dp, so that dP is a plane through the mean point of the reference tetrae- dron. Also, as dP is the limit of P P' as these planes ap- proach coincidence, it must always pass through the common line of P and P', and hence, at the limit, through the point of contact of P with the surface it envelops. 167. Tlie general homogeneous equation. We shall deal only with homogeneous equations ; for, by (377), 4p|e = 1, so that any term of lower degree than the highest can be raised to that degree by multiplying it by the proper power of 4j9|e, without changing thereby the meaning of the equation. Any homogeneous equation of the second degree in p may be written in the form 0, ........... (468) in which < is self-conjugate ; for such an equation can always be reduced to the sum of such terms as A^p\qi -p\q\\ that is, to the form 2,(Ap\q -p\q') = 0. This is equivalent to .p\q' + Aq' .p\q)] =p\$\_A(q .p\q' + q' .p\q)~\ if we write 208 DIRECTIONAL CALCULUS. [ART. 168. 168. Eq. (468) represents a surface of the second order; i.e. it is pierced in two points by any right line. The demonstra- tion is precisely as in Art. 105, and leads to the same value of y. given in (327). The equation > = .......... . (469) now represents a circumscribed cone with its vertex at q 1} while 3>|0e4p .... ........ (470) represents a circumscribed cylinder \\ to e. 169. Diametral planes. The equation of the locus of the middle points of a system of chords || to e is found, as in Art. 106, to be ............ (471) which represent now a diametral plane, conjugate in direction to e. Every line in this plane is also conjugate in direction to e. If p recede to oo, it becomes a vector, say e', and hence e'|p = 0. 170. Significance of the quantity \p. Differentiating (468) we have dp\p = 0; hence \p, which, by (468), is a plane through p, is the locus of the tangents to the surface at p ; i.e. the tangent plane. Its equation may be written ............ (472) Suppose \tj>p to pass through some fixed point e ; then we shall have e\p = = p\e. The point p in this equation is the point of contact of the tangent plane |e = 0, this equation is that of the locus of p, and is a plane, the polar plane of e. If e' be on the polar plane of e, we have e'|e', so that e is CHAP. VI.] SCALAE, POINT EQUATIONS. 209 also on the polar plane of e'. The points e and e' are conjugate to each other. We see, by (471), that a diametral plane is simply the polar plane of a point at GO ; hence the polar planes of all points in a diametral plane pass through the same point at co ; i.e. they are parallel to the diameter conjugate to this plane. If a point move along a line, its polar plane turns about a line. For let xp l + yp. 2 be any point on the line p^ ; then its polar plane is [0(opi + yp-2) = \ih + y'dPs, a plane through the common line of \^>p^ and typ? Also, since the polar planes of points on |^j and [

    2 P ass through Pip* If we put ^ and e 2 for p l and p% we see that the polar planes of points on a diameter pass through a common line at cc, i.e. they are parallel. 171. Center of the surface. This is the common point of any three diametral planes ; hence, if c^ *% c 3 are any points at oc, we have for the center q c = Jij^e^eo^cg, ......... (473) n being a scalar factor so taken as to make q t a unit point. To evaluate n, multiply both sides of (473) into 4 \e. /. 4 q c \e = 1 = _ (Ci< 2 (C3 (474) 4tyci4c*ta If we have ofa^esta, = 0, .......... (475) while the numerator of (474) is not zero, then the center is at oo, and (468) represents a non-central surface. If we put q c for q l in (469), we have the asymptotic cone ........... (476) 172. Sets of conjugate points. Any four points q^ q% q s , q which satisfy the six conditions, = 0, (477) 210 DIRECTIONAL CALCULUS. [ART. 173. form a set of conjugate points. Each point evidently lies on the polar plane of each of the others ; i.e. the four points form a tetraedron such that each vertex is the pole of the opposite face. There is an infinite number of such sets of points ; for, take any point in space as (ft; then any point in \(^q 1 as g 2 > by Art. 170, \(f>q 2 also passes through q ; next take any point in \qi(f>q 2 as q s ; then \(j>q. 3 passes through q^ 2 by Art. 170, and cuts \q 1 q 2 in g- 4 . If # 4 be at oo, qiq 2 q s is a diametral plane ; if q 3 be also at oo, then q^ 2 must be on a diameter, and, finally, if q 2 be at oo, q l must be the center of the surface ; thus q a e, e', e" are a set of conjugate points, three of which, being at oo, reduce to directions, as in Art. 169. 173. Normal set of conjugate points. If four points, besides the conditions (477), satisfy also the following fclfc = ?il& = & & = fclft = &!& = A = o, (478) they may be called a normal set of conjugate points. We pro- ceed to show that, with reference to any given surface jp|^*p=sO, there is one and only one normal system of conjugate points . 174. Solution of the equation p = np. Write the equation (< n)p = 0, and multiply its complement successively by any four points e, e', e", e'", thus obtaining e \((f} n)p=2^\(4 > ~~ n ) e = e 1 \(- ri)p = p($ n)e' = "!(*- n}p=p\(4>-n)e" =0 e'" (< - ii)p =p ( - n)e'" = Each of these equations must be satisfied by the same values of p that satisfy the given equation ; therefore p must be simultaneously on the four planes |(< n)e, etc. : hence these four planes must have a common point, the condition for which is (< - n)e( - n)e'( - n)e"(tf> - n)e'" = 0, (480) or i 4 k 3 n s -+- k 2 n 2 k^i + k = 0, CHAP. VI.] SCALAR POINT EQUATIONS. 211 in which the k's have the following values : i. (ee'eV'V (481) k 3 = (ee'e"e'")- l 2(ee'e"e"') j the summations being on the following plan, viz. : the <'s are to be placed before the different e's in succession, three by three, two by two, and one by one, for k 1} &, and k s , respec- tively; thus ee'ee'e"e"' + etc. The solution of (480) will give four values of n, which, sub- stituted in (479), will give the required points. Eq. (480) can have no imaginary root. This may be shown precisely as in Art. 112. Let the roots be n 1} n 2 , n 3 , n^ to which correspond the points Pi, P2, p s , p l ; then these points are given by the equations i- ni )e" = (n i-n 2 )e" = I, . . . (482) etc., etc. in accordance with (479). 175. The points p l5 p, just determined, constitute a normal set of conjugate points. Since <^p^ = n^p^, p 2 = n 2 p 2 , etc., we have 2h\P2 = n sPi\Pm etc. ; so that, ifp^pz = 0, then Pi\p 2 = also. As $ is self -conjugate, write it in the most general form of such a function in terms of %, n t and_pj, j? 4 ; i-e. Pi - n s p 3 ' p 2 \p In order that this form of < may satisfy the above condi- tions, we must have piPtptpt^lf Pr = Pr = P/ = Pf = 1> and Pi\P2=Pi\P3=Pi\Pi =Ps\P* =Pi\P 2 =P 2 \Ps = 0- These last con- ditions also imply \p 2 pap t = pi, etc., for they make p lie 212 DIRECTIONAL CALCULUS. [ART. 176. simultaneously on the three planes p 2 , |p 3 , |p 4 ; therefore Pi = m\p 2 p 3 pt, m being a scalar constant. Multiply into p v Pr = 1 = m\PzPsP^Pi - mpiP2PsP* =m, so that Pi= \p2PaP* Thus, whatever the original form of < may have been, by expressing it in terms of p^ 'p^ as determined in Art. 174, it will always be reduced to the form \PI + n 2 p, . p\p s + n s p 3 -p\p 3 + M 4 p 4 p\p t . (483) 176. Canonical form ofpl^p. With < as in (483), we have p\p=n l (p Pi) 2 +n 2 (p p^-}-n & ( < p\p & f+n^p\p^=^, (484) as a form to which p\(f>p = may always be reduced. This equation may also be written P (jhV*h +AV n 2 ) +P\ (PsVn~ 3 + J p 4 V- 4 ) or pfa -p\qj in which & J In this form it appears that the surface passes through the four common lines of two pairs of planes \q l} \qi and \q 2 , \q 2 . If these planes are all real, then real right lines lie wholly on the surface ; if any of them are imaginary, then there are no real right lines on the surface. 177. Rectilinear generators. Using q l} p, and is the criterion for distinguishing whether (468) represents a skew, developable, or convex surface. Jc is an invariant, as are also k 1; Jc 2 , and Jc s of (481) ; i.e. they are unchanged if any other points be substituted for e, e', etc. Now we have seen that any form of may be transformed by changing the reference points into the form (483) ; hence, as this transformation will not affect Jc , it, Jc Q , must have the same meaning for this form as for the original one. Eq. (483) gives & = n^n^ ........... (487) Considering positive and negative values of k as dependent on the signs of the w's, we have four cases : 1st. All the w's positive, k positive ; 2d. One of them zero, & = ; 3d. One of them negative, Jc negative ; 4th. Two of them negative, Jc positive. Three n's negative does not give a new case ; for, by chang- ing all the signs, we have only one negative. In the first case all the g's of (485) are imaginary, and no 214 DIRECTIONAL CALCULUS. [ART. 170. real value of p will satisfy (484), which therefore represents an imaginary surface. In the second case, suppose ?i 4 =0; then (48o) takes the form which represents a cone touching q l and \qj along their inter- sections with |j? 3 , imaginary if n l} n 2 , n 3 are all positive, real if one of them is negative ; thus we have a developable surface. In the third case suppose n 4 negative ; then g 2 and q. 2 ' are real, while q { and (?/ are still imaginary ; hence there are no real right lines on the surface, which is therefore convex. In the fourth case let n 2 and n 4 be negative. [It manifestly makes no difference which two are supposed negative, for any other pair as well as n 2 and n t might just as well have had negative signs under the radicals in (485).] We now have all the g's real, so that the surface has two real systems of recti- linear generators, and is therefore skew. If n s = w 4 = 0, the surface becomes simply two planes, real or imaginary, intersecting in a real right line. "We have thus ( positive for a skew surface ) Jc (= fye^e^e^e.^) < zero for a developable surface > . (488) ( negative for a convex surface ) In the value of A % we have put e , e 3 instead of e, e'", because the value of fc is unchanged thereby, and the reference points are the ones generally most convenient. 179. Nature of the surface at infinity. To ascertain this, let the variable point p recede to oo, by substituting for it P=o = P = in the equation of the surface p\p = 0. .-. (a*! + ye 2 + ze 3 ) ^(a*! + y s -f : 3 ) = or sft CHAP. VI.] SCALAR POINT EQUATIONS. 215 Eqs. (489) and (490) taken together represent a cone pass- ing through the intersection of p\p = with the plane at oo. If this cone be real, the surface has a real curve at oo ; if it be imaginary, the surface has no real curve at oo ; if it break up into two real or imaginary planes, the surface has two real or imaginary right lines at oo. Take a lt a^ a 3 as in Art. 140, so that 04 = [e^a, etc., this com- plement being regarded as referring to a vector system. Also write A, B, C, D, E, F for the coefficients in (490) taken in order, i.e. l 'p= 2 Ep t a s pa l -f 2Fpia 1 p a l = ) (401) - p ,, We will next ascertain for this equation the values of w , Mj, m 2 appearing in the table in Art. 164. These will be suffi- cient to determine the cone completely. We have, if we put X = q = e l e , etc., (Ea, + Daz+ Cog) A F E F B D EDO = (ej e n ) (e. 2 = 4 e p. Then we have (e 1 e ) = A& A^CQ, etc. Hence we find CHAP. VI.] SCALAR POINT EQUATIONS. 217 _ A m l = O m, = 3^o -f ^ + A 2 181. Consider next the equation 0, ....... (496) which represents a quadric surface passing through the four reference points, since it is satisfied when p = e , p = e^ etc. It is the most general equation of the second degree in p representing a locus passing through these points ; for it con- tains all the combinations of the quantities j)|e , p\e 1} etc., taken two at a time, and there must be no term containing only one of these quantities, as p\e , because this term would not vanish when p = e . Eq. (496) contains Jive arbitrary constants, and can there- fore be made to fulfil five conditions, such as to pass through five given points; but the locus already passes through the four reference points ; hence the general quadric can be sub- jected to nine conditions, and the general equation must con- tain nine arbitrary constants. We shall, in fact, obtain this general equation by adding together (495) and (496) . 182. Conditions in order that (496) shall represent a sphere. If (496) represent a sphere, the sections of the surface by the reference planes must be circles. Take the section by the plane p\e Q =pe l e 2 e s = 0, and the equation becomes A'p\e 2 'p\e s + B'p\e 3 -jp|ei + C"p| 1 -p\e 2 = 0. Consider the plane space fixed by e 1? e 2 , e 3 ; then, by (358), the condition that this equation shall represent a circle is 218 DIRECTIONAL CALCULUS. [ART. 183. AI fit rti = =2 in which a, S} =Te 2 e s , a 31 =7 7 e 3 e 1 , a l2 = Te^ 23 ^31 ^12 Proceeding in the same way with the other reference planes, we find the required conditions to be C 2 2 a 31 a^" (497) 183. EXERCISES. (1) Show that for eq. (496) we have the following values, in which 2% = B +C - A', 2W, = C + A-B', - vz.: KQ A B C A C" B' B C' A' C B 1 A' ~\ A A f^ 1 7? R' ^ C, .B' A, A B, ( t j > i AQ 33 m = - 8 G -B 21 3321 C m -4 ( ^S + ! B 2l , C33 > m 2 =-2(A + B+C). (2) Show that for the equation formed by adding (495) anc (496), we have *. ^ ^1 5 (7 ^ ^ C' B B C'A 2 A C B' A 1 A ^o, ]> 62, 6 3 -^1 ^^ -^IG? -^-i -^-? ^ "*~ -^? -^ "~ ^ ' ' ^c == R xl /^' A A Tt A 1 f 1 -D -*lo) ^ -**J -^*2 "> " ^ j C AQ, B A) A B, A 3 C .,- 21' G" 33" sg" 2(" g' 21' G" , 33' 21" , G' 33" 1 ~~ G" 33' 21" G' 33" 21' ' in which S" = A U + B'-C-A, (5" = A () +C'-A- B. CHAP. VI.] SCALAR POINT EQUATIONS. 219 (3) Show that (496), with the conditions (497), represents a sphere, by transforming the equation to a vector system. See Art. 75. (4) Show that the most general equation of the second degree in p, in solid space, may be written p\e p\e ' -f-p|ei -p\e^ -\-p\e% -p\^ -\-p\e$ 2 ) \ e s == ^ 5 (498) write the corresponding self-conjugate form of <, and deter- mine p = 0, determining thus the equation of the tangent plane independent of the point of contact. By using the result, find the tangent plane conjugate in direction to c. Ans. 4 Ce\~*e=e c^^ei v' ee\(f>~*eff>~ l e, 4p[ 2 P = 0, show that the locus of its pole, with reference to any other quadric : 0, iS (J\(l).^ l (t>, 2 Q =0. 220 DIRECTIONAL CALCULUS. [ART. 18,. (11) Find the nature of the surface in each of the following cases, the position of the center, and the asymptotic cone, if real : (a) In (496) let A = B = O= A' = B' = C' = 1. Ellipsoid, (b) In (496) let C" = 0. Skew or developable surface. (c) In (496) let ' = C" = 0. Skew surface. (d) In (496) let A = B = C= 0. Cone, vertex at e^ (12) In cases (&), (c), (d) of the last exercise, show that one, two, and three edges respectively of the reference tetrae- dron lie on the surface. (13) Show how the cone of Ex. (11), case (d), is related to the surface A <) (p\e (> y + A'p\e 2 jp|e 8 + B'p\e s -p\e^ + C"p|e 1 -p]e 2 = 0. (14) Discuss, as in Ex. (11), the following cases : (a) In (496) let A = B = C = 1, A' = B' = C' = 2. Ellip- soid. (6) In (496) \etA=B=C=A' = B' = l,C'=3. Elliptic paraboloid. (c) In (496) letA = -B = C=-A'= B'= - C'= 1. Two- sheeted hyperboloid. (d) In (496) let A = B = C = A' = B' = 1, C' = 0. Elliptic cylinder. (e) In (496) let A = - B = C = - A' = B 1 , C' = 0. Cone, vertex at i(e 2 -f e 3 ). (/) In (495) let A = A 1 = A 2 = -A s = l. Two-sheeted hyperboloid. (g) In (495) let Aq = A l = A 2 = A 3 . Hyperbolic para- boloid. (/*) In (495) let A = 2, A, = A 2 = 4, A 3 = - 1. Elliptic paraboloid. CHAP. VI.] SCALAR POINT EQUATIONS. 221 (t) In (498) let e ' = ^(e l + e a + e s ), ei' = i(e 2 + e 3 + c ), **' = i(s + e o + e i) e -3 = K*n + i + e,) . (j) In (498) let p = and p\<{>p = C are similar, concentric, and similarly placed. (20) Discuss the following equations : 222 DIRECTIONAL CALCULUS. [ART. 184. 2 - ,,)= - = e 2 ,) ? (21) Find the equation of the surface reciprocal to p\p=Q; i.e. the locus of q = p. 184. Inversion of <. If we have given

    in order to find p as an explicit function of q. Taking, for generality, < as not self-conjugate, let p=p c q. Let e, e' } e", e'" be any four points ; then we have e\p =p c e = e\q, e'\$p = p^ c e' = e'\q, e" p = p\ c e" = e"\q, e'"\p = p&e'" = e'"\q. Now substitute in (201) (f> c e for jp , c e' for p 1} etc. = ,., -|^X^" e e" > . g ie + IV>ce">,g-g|e'-eto. ( ^ 9} c e c e' c e" c e"' If we put q = [e'e"e" f , which we can do whatever point q may be, because the e's are any points whatever, and hence three of them may always be so taken that the complement of their product shall be q, then (499) reduces to Jc - l \e'e"e'" = l&e'&e'tyX", ..... (500) Jc having the value given in (481) except that < c must be put for <. Now substitute + n for , where n is any scalar. On expanding the last member, the first term is k ~ l q, the last is n s q, and, as the first member is a function of q, it ap- pears that the coefficients of 7; and n- should also be functions CHAP. VI.] SCALAR POINT EQUATIONS. 223 of q. Call them \q and \j/q respectively ; then, expanding & ', the equation becomes Operate on this equation by < -f- n. .-. (n* + k 3 n 3 -+ Ay i 2 + k^i + ) q This equation must hold for all values of n ; hence the co- efficients of the same powers of n on each side of the sign of equality must be equal. .-. \lr = A'.., <^>, and x = k k s (f> + q -}- Ay-g - *q, ) ^ or <>* - A 3<^ 3 + A- s ^ 2 - A-,^ + A- = 0. )' Substituting the values of if/ and x i n the value of ( + we obtain also (dl i n ^i rj = ^~ 1 g+(^2- ^ 185. Equation of the anti-polar plane. This is found pre- cisely as in Art. 121, and is of the same form, viz. : i>|*(2g e -0 = 0, ....... (503) e being the point of which this equation represents the anti- polar plane. 186. Reciprocating ellipsoid. To obtain the equation of this surface we proceed in the same manner as in Art. 122 for the reciprocating ellipse. Assume the equation P\P =P\ e o ' P ( e i + ^ + (2e e) = 0, or, putting e for e, and its value for e, j9|(ei + 6 2 + e 3 e ) = 0. We must then have |e coincident with \ [8 me + e 2 + e 3 + 6 + etc.] = e [16 me +6j + e 2 + e 3 ] = (4m This expression becomes zero if m = ^, which value, on substitution, gives for the required equation - P\ (e! + + (e 2 + e s + e d) p e^ + etc. > * 187. We will now show that the complement of any point is its anti-polar plane with reference to the ellipsoid of (504.) If this is true, we must have \p \(2e p) = or p(2e -p} = 0. Let p = 2 we, with the condition S n = 1 ; then p(2ep) = 2we <[( n )e + (| n l )e l + etc.] [ (i w ) (ei + e 2 + c 3 e o) + etc.] [(1 ?j w, 7? 3 + ?j )e + etc.] e = 2j = 0. Q.E.I>. CHAI-. VI.] SCALAR POINT EQUATIONS. 225 188. Scalar f)lane equations. In eq. (496) put P for \p, and we obtain the complementary or anti-polar reciprocal equation Pe . P(Ae, + Be, + Ce z } + Pe, - P(C'e 2 + B'e 3 ) ) e 2 .Pe s = P ' ( 6) which is that of a surface touching the four reference planes, since it is satisfied when P = |e , P= \e 1} etc. If we write > , (507) + (A'e s +Be + C'e,) Pe r t(Ce 9 +ye l +A l e t ) Pe 3 1 eq. (506) becomes P|^P=0, (508) and i/r is a linear, self-conjugate function of P. In the same way any scalar point equation p\p = may be changed into its complementary plane equation Pi/rP=0. Bearing in mind that, if P= \p, then |P=|(|p) = p, we have the following relations between < and ^, viz. : (509) Suppose (508) to represent any homogeneous, plane equa- tion of the second degree. The equation shows at once that the point \\j/P lies on the plane P. Differentiating, we have but, by Art. 166, dP is a plane through the point of contact of P with the surface. Hence \\{/P is on the line PeZP. Now dP is a variable plane, subject only to the conditions of always passing through e and the point of contact of P; hence \\j/P must coincide with the point of contact of P. If Q be any plane not tangent to (508), then \\j/Q is its pole with reference to that surface. 189. Center of surface P|^P=0. The center is the pole of the plane at oo, i.e. of \e; hence, by (509), q e = m\\jr\e = m\ (|e 4 e = 4me|6 whence ^ = TTT^ ( 51 ) If we have e\e = 0, (511) the center is at oo, and the surface is a paraboloid. This is also the condition that the reciprocal surface p\p = shall pass through the mean point of the reference tetraedron. 190. Reciprocal surfaces. A skew surface is reciprocal to a skew surface. For such a surface is generated by a right line whose consecutive positions are not coplanar ; hence the recip- rocal surface is generated by a right line whose consecutive positions have no common point. A developable surface is reciprocal to a curve, and vice versa. For such a surface may be regarded as the envelope of a plane rolling on some two given surfaces Si and S 2 ; hence the point reciprocal to this plane lies simultaneously on the two sur- faces reciprocal to S t and S 2 ', i.e. it generates their common line. When the developable surface is a cone, the reciprocal curve \splane; for, since all tangent planes to the cone have a common point, their reciprocal points are coplanar. A convex surface is reciprocal to a convex surface. This fol- lows from the preceding, since all surfaces are included under these three heads. 191. The discriminant k as a criterion. Since, by (509), tf/\ p = |P = is a-^ plane curve ( convex surface ) 192. To determine still farther the surface represented by (508) we will make use of the cone circumscribed about the surface p\p = and having its vertex at e, the center of re- ciprocation. If this cone is real, the surface cuts the plane at CHAP. VI.] SCALAIJ POINT EQUATIONS. 227 oc in a real curve, the reciprocal of this cone. If the cone is imaginary, the surface has no real points at cc. Substituting e for q t in (469), we have the equation of the above-mentioned cone, viz. : pe\e = () (512) Write 'p = p e\(f>e <$>e p\e, . . . . . (51o) and with this value of <' compute M O , m^ m. 2 according to Art. 179, and determine by the table in the same article whether (;")!-) is real or imaginary. We obtain thus the following scheme for the determination of the equation P\j/P=. SURFACE. A) CONE OF EQ. (512). Ellipsoid. Imaginary. Ellipse. (i Imaginary ellipsoid. + " Elliptic paraboloid. - Two coincident planes tangent at e. (e\ and if/, p and P, are related as in Art. 188, show that p\$"*p = and P\\I/P=Q are respectively the point and plane equations of the same surface, reciprocal to that represented by jp| = 0. Also that 7 J i/r" 1 P=0 and p\<}>p = are respectively the plane and point equations of a surface reciprocal to that represented by P\\j/P= 0. CHAP. VTL] APPLICATIONS TO STATICS. 237 CHAPTER VII. APPLICATIONS TO STATICS. 194. It is proposed, in this concluding chapter, to give a few applications of our calculus to mechanics, merely to serve as an introduction to this field, and to indicate how perfectly the methods that have been developed adapt themselves to mechanical conceptions and processes. 195. A force is that which is postulated as the cause of any change, or tendency to change, in the rate of motion of some particle, or rigid body, on which it acts. In statics we consider systems of forces aside from any ques- tion of rest or motion of the bodies on which they act ; and, especially, cases in which the total resultant effect of all the forces applied to a body is null. 196. The space qualities of a force are magnitude, direction and position, and these are the only qualities with which we have to do mathematically. The intrinsic character of a force, such as that of gravity or magnetism, we know little or nothing about; but our knowledge is complete for its mathematical treatment, when we know its magnitude, direction, and line of action, or position. Xow these space qualities are identical with those of a point-vector ; hence, for the purposes of mathe- matical discussion, a point-vector completely represents a force, and therefore all that has been demonstrated in Chapter II. regarding the former can be applied immediately to the latter. 197. Notation. The notation for points and vectors, in gen- eral, will be as in previous chapters. The vector representing the magnitude and direction of a force will be denoted by a 288 DIRECTIONAL CALCULUS. [ART. 198. German letter as ^, while the magnitude of the same force will be denoted by F, the corresponding English letter. Thus F will be the tensor of ^, or F= T$. If e be a point on the line of action of the force, it will then be completely denoted by e$, a notation which is practically more convenient than the use of a single letter to represent the point-vector or force. The complement will be used in this chapter only with refer- ence to a vector system in solid space. 198. Forces acting on a particle. The parallelogram and polygon of forces follow at once from the nature and properties of vectors and point-vectors, as shown in Chapters I. and II. Let a system of forces acting on the point e be denoted by e #i> e ?$2) e &n > then the resultant effect of the system will be found in this, as in all cases, by simply adding the forces. Thus Resultant = 2eJ = For equilibrium we must have eSJ = 0, or 2g = 0, ............. (515) an equivalent equation. 199. Equilibrium of a particle contained to remain on a smooth curve. In this case the resultant force must have no compo- nent along the tangent to the curve, at the point where the particle is ; hence the resultant must be _L to this tangent. Let the equation of the curve be p e 9 =p = t, being a vector function of the scalar variable t ; then & = *? = & at dt is a vector along the tangent. Hence, if p be the position of the particle on the curve, the condition for equilibrium is ...... (516) CHAP. A T TL] APPLICATIONS TO STATICS. 239 For example, if the curve become the right line whose equa- tion is p = e + e't, then dp = t'dt, and the condition becomes e'|S3f = 0. Again consider the case of a particle resting on a diagonal of a parallelepiped, and acted on by three forces represented by the edges of the parallelepiped which meet at a corner not on the diagonal. Let the three edges be cj, e>, e 3 , and the equation of the diaonal thus 25 = c 1 + e 2 + e3 ? and = so that we have or ( e2 + 3 )_ l = ; i. e . ei =e 2 + e 3 ). 200. Equilibrium o/ particle constrained to remain on a smooth surface. If v be a vector || to the normal at p, then for equilibrium 2^ must be || to this vector. .-. vS$ = ............ (517) is the required condition. If the equation of the surface is a scalar one in terms of vectors, then it will be linear in dp after differentiation and will have the form v\dp = 0, v being some function of p. To illustrate, let the equation of a surface be then 3 (p^) 2 . dpfa + 2 P [e 2 . dp\e 2 - d p \t s = 0, or d p \(3 fl (p^) 2 + 2*,. p[e 2 - 3 ) = rfp'v = 0. If we have a vector equation, it will be in the form p = <(, y). Then - and -- will be vectors II to tangents to the surface at dx dy the end of p, and hence we may write (518 > 240 DIRECTIONAL CALCULUS. [ART. 201. 201. EXAMPLES. (1) If e 1} e 2 , ^3 are the vertices of a tri- angle, and p l} p< 2 , p s the middle points of its sides, p l opposite gj, etc. ; then forces represented by e^p^ e 2 p. 2 , e 3 p s are in equi- librium. We have = \ (ei s + ei*8 + 6263 + Cai + e 3 e, + e,e 2 ) = 0. (2) Forces are represented by perpendiculars drawn from the vertices of a triangle on the opposite sides ; to show that they cannot be in equilibrium unless the triangle is equilateral. Let the triangle be e^s ; a = Tie 2 e 3 , b = Te s e^ c = Tefa ; p^ foot of _L on e.,e s , etc. ; I, m, n cosines of angles at e 1? e.,, e s . Then, p l = - (bne 2 + cme s ), p 2 = - (cle 3 + awe x ) , p s = - (ame 1 + ble. 2 ) , and +Mi Ws+f - - \6 c/ \c a which cannot be zero unless a = 6 = c. (3) Let ej, e 2 j e s be any three unit vectors, and e^F^ ee.,F. 2 , f s F 3 three forces acting at e. Then for equilibrium 2g = 2cF= e^ + 2 F, + e 3 F 3 = 0. This gives e l e^e 3 = 0, so that the forces must be coplanar. Also, But the three vectors, being coplaiiar, may be taken as in plane space, and, therefore, ej 2 , etc., scalar ; so that e^., = sin < c -. e i etc., and we have (4) If EI, e,, 3 , 4 be any four unit vectors, and ee^, ee.yF.,, etc., be four forces ; show that for equilibrium we have CHAP. VII.] APPLICATIONS TO STATICS. 241 (5) Show that the point on the smooth surface where a particle attracted toward the origin will remain at rest, is given by a 3 W c 3 ^/ a 6 + (6) Through a point at the end of c three chords are drawn, parallel to a set of conjugate diameters of a central quadric ; forces act on this point, represented by the portions of the chords intercepted between the point and the surface, and towards the surface : show that 2# = 2e. Let a, /?, y be conjugate semi-diameters, and equations of chords p = e + xa, p = e-f y(3, p = t + zy. Substitute the first value of p in the equation of the surface ; .'. (c + Ofa)|4(c + Xa) = 1 = |0 + y /a efiy + /3 . eya + y * ~ The last results are by eqs. (423) and (177). (7) Show that, if a system of forces acting on a point are represented by vectors drawn outward from the point, and are in equilibrium ; then this point is the mean of the extremities of the vectors. 242 DIRECTIONAL CALCULUS. [ART. 202 202. Forces acting on a rigid body. Let the forces be e^, e 2$% e ^c. ; then we have for the total resultant effect 28 = $e% = e^fi + 2 (e - e ) g = e 2$ + Sefr (519) In the last member we have written e x for e,^ e , etc. Comparing with Art. 61, we see that the quantity 28 has the same space qualities as the quantity there denoted by S, and called a screw. Following again the nomenclature of R. S. Ball, Astronomer Eoyal of Ireland, we shall call the quantity 28 a wrench, and it is evident that it corresponds to a screw, just as a force does to a point-vector. Before proceeding to a general discussion of eq. (519), we will consider some special cases. 203. Parallel forces. Let ^ 1 =F 1 , % 2 =F 2 e, etc., and Te=l; then 28 = 2e$ = *ZeF which, by (519), reduces 2B to a single force ; and, by 2$ = 0, which reduces the wrench to a couple, or zero force at oo. It follows that any system of forces confined to one plane will be equivalent either to a single force, or to a couple. 206. For equilibrium we must have which, because 2S is the sum of a point-vector and a plane- vector, requires that we have S8 =0 1- = 0) < 526 > 207. Normal form of a wrench. It appears, by Art. 67, that, by properly choosing the point at which the resultant force acts, !$ may always be reduced to the sum of a force at q, and a couple whose plane is perpendicular to the force. 244 DIRECTIONAL CALCULUS. [Ant. 208. Writing in (215) 2$ for a, and 2eg for ]/?, we have . (527) and, by (214), the vector perpendicular between e 2$ and q^ is (528) in which p = q e . By Art. 46, the second term of the third member of (527) is the orthogonal projection of the resultant couple SeJ on a plane _L to 2$, i.e. on |2$. Now the orthogonal projection of any plane-vector upon a plane is always less than the projected plane-vector in magnitude; hence the couple in the normal form of 2B is the minimum of all the couples obtained, when e n occupies positions not in the line of q^, which is called the axis of the wrench. 208. Recurring to eq. (213), let us consider the equation T(\p-pa)=C, ...... ...... (529) i.e. the area of plane-vector part of S constant. Taking the co-square, we have j8S - 2pa/3 + (pa) 2 - = p-2pa(3 + pW - (pja) 2 = G\ or P ^-(p\aY-2pal3=C 2 -^- ........ (530) Writing tf>p = a-p a p|a, so that the first two terms be- come p\p, and comparing with (443), we have y = - \ a p, and (C of (443) ) = C 2 - &. . . . (531) In (449) put X = a, /x = /3, v= |a/3, so that |a/3 = a? a/3, and we see that 8 = -, so that the surface represented by (529) CHAP. VII.] APPLICATIONS TO STATICS. 245 is a cylinder. We have, as in Art. 160, for the equation of the axis of this cylinder 8 + y --- ; i.e. a?S - a 8\a - \a/3 = 0, or, by (189), aS,a = aft .......... (532) which is identical with (214), from which (528) was derived. Thus the axis of the cylinder coincides with the axis of the screw, qa. The discriminating cubic (461) becomes (g a-) 2 = 0, so that the cylinder is circular. Now putting, as in the last Article, 2$ for a and Se^y f r I A we see that the locus of q, tchen the moment of the resultant couple is constant, is a circular cylinder tvhose axis is the axis of the wrench. 209. Any wrench can be reduced in an infinite number of ways to the sum of two forces. Let us write 5B = TF(ee + a|e), to which form we have seen that any wrench is reducible. Now put Wa\f. = (e' e)#, which requires e' e and g- to be both JL to e, but can be satisfied by an infinite number of values of e' and $, subject to these conditions. Substituting, we have 2S = Wee + (e 1 - e) g = e( We - which, by what we have just seen, proves the proposition. The tetraedron of which the opposite edges are any two forces whose sum is 2B is of constant volume. Let p^ and ;> 2 ^2 be two such forces, so that 2h% +^2^2 = 2B- Therefore, squaring, The first member of this equation is 12 times the volume of the before-mentioned tetraedron, and the last member is a con- stant when 2S is given, which proves the proposition. 210. EXERCISES. (1) Find the conditions in order that the orthogonal projections on any direction whatever of a system of forces may be themselves a system of forces in equilibrium. 246 DIRECTIONAL CALCULUS. [ART. 210. Let t be a unit vector in any direction; then the projection of $! on t is i i|^!, and similarly for the other forces. By (526) we have for equilibrium, putting t t]^ for ^ etc., As this is to be true for all directions of t we must have 25 = 0. Again, Avriting t t|J for g in the second of (526), we have S(cHf)--*3(e.*|8r)0 as the other condition, which requires 2(e i $) to be || to t. (2) Forces act at the vertices of a tetraedron, in directions respectively _L to the opposite faces, and proportional to the areas of these faces in magnitude ; shoAv that the forces have the property discussed in Ex. (1). Let e^, e e 2 , e e 2 be three edges of the tetraedron ; then, Hence we have at once 2^ = 0. Also, so that the second condition is fulfilled. (3) A cube is acted on by four forces ; one is in a diagonal, and the others in edges, no two of which are coplanar, and which do not meet the diagonal ; find the condition that they may be equivalent to a single force. Ans. If $!, ^ 2 , $ 3 are along the edges, and g 4 along the diag- onal, the condition is F t (F l + F,+ F 8 ) VJ + F,F 3 + F S F, + F,F, = 0. (4) Six equal forces act along six successive edges of a cube which do not meet a given diagonal ; find the resultant wrench. Ans. If F be the magnitude of each force, and the edges of the cube be e ij, e^, e i 3 , we have (5) If 2B = efa + e 2 $ 2 , and & is _L to g, and $ 2 , find the normal form of 28 ail( i the position of its axis. CHAP. VII.] APPLICATIONS TO STATICS. 247 Let e % (G! + e 2 ) , and e = e 2 e = (e l e n ) . Then ISi= i#2= Hence Vtfi T ffj By (528) the projection of g e on a plane _L to 2$ is Ff-F? so that, if we take g e _L to 2$, we have (6) Three forces whose magnitudes are 1, 2, and 3 act along three successive edges of a unit cube which are not coplanar ; show that the vector equation of the axis of the wrench is TVs -M( tl + 2i 2 + 3t 3 ). (7) Forces act at the mean points of the faces of a tetrae- dron, _L and proportional to the faces on which they act ; show that they are in equilibrium. (8) Let p be any point within a tetraedron, and let a system of || forces act at the vertices, each proportional in magnitude to the tetraedron formed by joining p with the other three vertices ; find the centre of || forces. (9) The sides of a rigid plane polygon are acted on by forces _L to the sides and proportional to them in magnitude, all the forces acting in the plane of the polygon, and being directed inwards ; also the sides taken in the same order are severally divided by the points of application in the constant ratio of m to n ; show that the system of forces is equivalent to a couple whose moment is /H m ~ n ) ^ j n w hich u. is the 2(m + n) ratio of the magnitude of any force to the length of the side it acts upon. MATHEMATICS. 81 The Method of Least Squares. With Numerical Examples of its Application. By GEORGE C. COM- STOCK, Professor of Astronomy in the University of Wisconsin, and Director of the Washburn Observatory. 8vo. Cloth, viii + (i8 pages. 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PT1HE peculiarities of this treatise are the rigorous use of the Doctrine of Limits, as a foundation of the subject, and as preliminary to the adoption of the more direct and practically con- 82 MATHEMATICS. venient infinitesimal notation and nomenclature ; the early intro- duction of a few simple formulas and methods for integrating ; a rather elaborate treatment of the use of infinitesimals in pure geometry ; and the attempt to excite and keep up the interest of the student by bringing in throughout the whole book, and not merely at the end, numerous applications to practical problems in geometry and mechanics. Elements of the Integral Calculus. Second Edition, revised and enlarged. By W. E. BYERLY, Professor of Mathematics in Harvard University. 8vo. xvi + 383 pages. Mailing Price, $2.15; for introduction, $2.00. 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Vaughn, Prof, of Mathe- matics, VanderMlt University: It is pleasing to see the author avoiding, and in some cases leaving out of sight, the old ruts long since worn smooth by our teaching fathers. MATHEMATICS. 83 A Short Table of Integrals. Revised and Enlarged Edition. To accompany Byerly's Integral Cal- culus. By B. O. PEIRCE, Jr., Instructor in Mathematics, Harvard Uni- versity. 32 pages. Mailing Price, 15 cents. Bound also with the Calculus. Byerly's Syllabi. By W. E. BYERLY, Professor of Mathematics in Harvard University. Each, 8 or 12 pages, 10 cents. The series includes, Syllabus of a Course in Plane Trigonometry. Syllabus of a Course in Plane Analytical Geometry. Syllabus of a Course in Plane Analytic Geometry. (Advanced Course.') Syllabus of a Course in Analytical Geometry of Three Dimensions. Syllabus of a Course on Modern Methods in Analytic Geometry. Syllabus of a Course in the Theory of Equations. Elements of the Differential and Integral Calculus. With Examples and Applications. By J. M. TAYLOR, Professor of Mathematics in Madison University. 8vo. Cloth. 249 pages. Mailing Price, $1.95 ; Introduction Price, $1.80. rFIIE aim of this treatise is to present simply and concisely the fundamental problems of the Calculus, their solution, and more common applications. Many theorems are proved both by the method of rates and that of limits, and thus each is made to throw light upon the other. The chapter on differentiation is followed by one on direct integra- tion and its more important applications. Throughout the work there are numerous practical problems in Geometry and Mechanics, which serve to exhibit the power and use of the science, and to excite and keep alive the interest of the student. The Nation, New York: It has two marked characteristics. In the first place, it is evidently a most carefully written book. . . . We are acquainted with no text-book of the calculus which compresses so much matter into so few pages, and at the same time leaves the impression that all that is necessary has been said. In the second place, the number of carefully selected examples, both of those worked out in full in illustra- tion of the text, and of those left for the student to work out for him- self, is extraordinary. Mathematics. 2 Introd. Prices, Byerly , Differential Calculus $2.00 Integral Calculus 2.00 Ginn. .... Addition Manual if Halsted Mensuration 1.0* Hardy Quaternions 2.00 Hll 1 . Geometry for Beginners 1.00 Sprague Kapid Addition 10 Taylor Elements of the Calculus 1.80 Wentworth Grammar School Arithmetic .75 Shorter Course in Algebra 1.00 Elements of Algebra 1.12 Complete Algebra 1.40 Plane Geometry 75 Plane and Solid Geometry 1.25 Plane and Solid Geometry, and Trigonometry 1.40 Plane Trigonometry and Tables. Paper. . .60 PI. and Sph. Trig., Surv., and Navigation . 1.12 PL and Sph. Trig., Surv., and Tables 1.25 Trigonometric Formulas 1.00 Wentworth & Hill : Practical Arithmetic 1.00 Abridged Practical Arithmetic 75 Exercises in Arithmetic Part I. Exercise Manual Part II. Examination Manual 35 Answers (to both Parts) 25 Exercises in Algebra 70 Part I. Exercise Manual 35 Part II. Examination Manual 35 Answers (to both Parts) 25 Exercises in Geometry 70 Five-place Log. and Trig. Tables (7 Tables) .50 Five-place Log. and Trig. Tables ( Comp. Ed.) 1.00 Wentworth & Reed : First Steps in Number, Pupils' Edition .30 Teachers' Edition, complete .90 Parts I., II., and III. (separate), each .30 Wheeler Plane and Spherical Trig, and Tables 1.00 Copies sent to Teachers for examination, with a view to Introduction on receipt of Introduction Price. GINN & COMPANY, Publishers. BOSTON. NEW YORK. CHICAGO. UCLA ACCESS SERV.CES Interlibrary Loan 11 630 Universitl Researcr Box 951 575 I Los Angeles. CA 9009b 1575 Engineeririf & Mathematical Scrences Library A 000169708 5 JUL72 SOUTHERN BRANCH, UNIVERSITY OF CALIFORNIA, LIBRARY, OS ANGELAS. CALIF.