519] 
ON CURVATURE AND ORTHOGONAL SURFACES. 
297 
We have the following identities: 
(a) X + (li) Y + (g ) Z — 0, 
(h)X + (b) Y + (f)Z= 0, 
(9)X + (f)Y+(c) Z = 0, 
((b)(c)-(fy,.., (g) (h) - (a) (/),...)=- (X s , Y\ Z\ YZ, ZX, XY)$, 
that is, (b) (c) - (/) 2 = - X 2 cf) &c., where 
= (be —f 2 , ..gh — af,. .$X, Y, Z)\ 
Writing also 
ciX + h Y+gZ, hX + b Y + fZ, gX -\- fY -\- cZ — 8X, 8 Y, 8Z, 
and X 2 + Y 2 + Z- = V 2 ; also a+b + c — w, then 
(a) = (b + c) V 2 - coX 2 + X8X - Y8Y-Z8Z, 
(b) = (c + a) V 2 - o)Y 2 - X8X + Y8Y-Z8Z, 
(c) = (a + b)V 2 - a>Z 2 - X8X -Y8Y+ Z8Z, 
(/) = - fV 2 -coYZ + Y8Z + Z8Y, 
(g) = -gV 2 - coZX + Z8X + X8Z, 
(h) = - hV 2 - «IF + X8Y + Y8X. 
14. I give also the following lemma: 
Lemma. The condition in order that the plane X£+Yrj + Z%= 0 may meet the 
cones 
(A, B, G, F, G, v , Z) 2 = 0, 
(A', B' } C', F', G', H'%1 v , £) 2 = 0 
in two pairs of lines harmonically related to each other, is 
(BG' + B'G - 2FF',.., GH' + G’H-AF'-A'F,..\X, Y, Z) 2 = 0. 
Writing here 
(A,.JY£-Z V , ZI-X& Xi) — Y%) 2 
= ((A), (B\ (0), (F), ((?), (H№, n, ?)=, 
that is, (A) = BZ 2 + GY 2 — 2FYZ, &c., the condition may be written 
(A)A' + (B)B' + (C) C' + 2(F)F' + 2(G)G'+2(H) H = 0, 
or say 
((A ),..\A= 0; 
and we may, it is clear, interchange the accented and unaccented letters respectively. 
c. viii. 38