236 PROFESSOR CAYLEY OX CURVATURE AND ORTHOGONAL SURFACES, 
whence 
(A,..OCo,...)=0. 
18. By what precedes, we have 
((A),..0& 9, £j 2 =0 
for the equation of the two principal planes, where the coefficients (A), (B), &c. are 
functions of A, B, &c. and of X, Y, Z, as mentioned above. These coefficients satisfy 
of course the several relations similar to those satisfied by (a), (b), See., and other rela- 
tions dependent on the expressions of A, B, &c. in terms of a, b , &c. and X, Y, Z. 
19. Proceeding to consider the coefficients (A), (B), &c., we have then 
(A)+(B)+(C) = (A+B + C)V 2 -(A, ..XX, Y, Z) 2 , 
that is 
(A)+(B) + (C)=0. 
Observing the relation A + B + C = 0, the equations analogous to 
( a )=(b+c)\ 2 -(a+b + c)X 2 + & c., are (A)= -AV 2 +X&'X-YS'Y-ZS'Z, &c. 
if for a moment we write b'X, b'Y, b'Z to denote the functions 
AX+HY + GZ, HX+BY+FZ, GX+FY-j-CZ. 
But, from the above values, Xci'X + Yb'Y -f- Z^'Z = 0, or the equation is (A) = — AY 3 + 2Xc> / X, 
that is = — AY 2 + 2X(Z§Y — Y!$Z). The equation for (F) is (F)= — FV 2 + Yd'Z-f-Z&'Y, 
where YS'Z+Z&'Y is =Y(Y&X-X&Y)+ Z(X$Z-Z&X), viz. this is 
=(Y 2 — Z 2 )BX— XY^Y+XZ^Z. 
We have thus the system of equations 
(A) = — AV 2 . +2XZoY — 2XY&Z, 
(B) = — BY 2 — 2YZ£X . +2XYoZ, 
(C) = -CV 2 +2YZSX -2XZW 
(F) = — FV 2 + (Y 2 — Z 2 )£X — XYhY + XZM, 
(G) =— GY 2 + XYSX + (Z 2 — X 2 )SY - YZdZ, 
(H) = — FIV 2 — XZ£X +Y72Y -f(X 2 -Y 2 )SZ. 
20. Wc hence find 
( A)a + (IT) h + ( G )g = - (A a + HA + Gg)Y 2 + (ZSY - Y£Z )oX + XP, 
(H)J+(B)fl + (Fj/= — (I1A + Y>b +F/)Y 2 +(XSZ-ZSX)W+YQ, 
( G )g + (F )f -j- ( C) c = -{Gg +F / +Cc )Y 2 +(Y^X~XhY)lZ +ZR, 
if, for shortness, 
P = ( g Y - h Z)IX + (aZ -gX)lY + (hX - aYfiZ, 
Q=(/Y — bZ)lX + (liZ —fX)lY + {bX - hY )&Z, 
E= ( cY -/ZJSX + (gZ - cX)<SY+(/X- yY)!SZ. 
