1036 
Proceedings of Royal Society of Edinburgh . [sess. 
10. Direction Cosines of the Transverse Component of Force . — It 
is desirable to find expressions for a sin 0, b sin 6, c sin 0, where 
a, b, c are the direction cosines of the transverse component. We 
have at once 
aa + b/3 + cy — 0 . 
Also, if l, m, n be the direction cosines of the normal to the 
plane containing (a, /I, y) and (X, g, v), we have 
la + mb + nc — 0 . 
But la + m/3 + ny = 0, and l\ + mg + nv = 0, so that Igm :n — 
/3v — yg : yX — av : ag — /3X. Therefore 
a(/3v — yg) + b(yX — av) + c{ag — /3X) = 0 , 
and we get 
a :b : c =* (3{gg — /3X) — y(yX - av):y((3v - yg) - a(ag - /3X) :a(yA - av) — /3(/3v — yg) 
= a(aA + /3g + yv) — X : (3(aX + /3g + yv) — g : y(aA + /3g + yv) — v . 
Writing a — &[a(aA + j3g + yv) - A], etc., we find 
Xa + gb + vc — A[(aA + /3g + yv ) 2 — (X 2 + g 2 + v 2 )] 
= - k( X 2 + g 2 + v 2 ) sin 2 0 . 
But Xa + gb + vc = (X 2 + g 2 + v 2 ) 1 sin 0 , 
so that k = - 1/(A 2 + g 2 + v 2 )* sin 6 . 
Hence (X 2 + g 2 + v 2 fa sin 0 — X — a(aX + /3g + yv) , 
the other two expressions being got by cyclical interchange of 
a, b, c; a, (3, y; X, g, v. 
11. Components of the Transverse Force . — The components of 
the first term in the expression for the transverse force at the 
origin (§6) are 
2* sin 0 cos 0 = + f3g + yv)[A - a(aX + fig + y^)](X 2 + g 2 + v 2 )~ B/2 , 
r 6 p 6 
and two other quantities obtained by cyclical interchange of a, f3,y ; 
X, g, v. At the positive pole, distant a forwards in the direction 
(a, (3, y), this becomes 
6J2M 
aX + f3g + yv+ 
X + J2 a— — a( aX + f3g + yv + 
■4 
(X 2 +a 2 + v 2 yv(l + 2 J2 ~y, 
a aX 4- (3g 4- yv 
+ g 2 + v 2 
+ 2 
a* 
X 2 + 
i y /2 
g 2 + v 2 ) 
