of Edinburgh , Session 1881 - 82 . 
347 
the two components, the one p 0 perpendicular to y, and the other p x 
parallel to y x , so that, if p 1 = 2y 1 , we have 
P = Po + zyi, 
z being a scalar. 
Taking of both members the function £, we have 
ip = ip 0 + • 
First we remark that, as <fry 1 — 0, we have also <^> 2 y 1 = 0. Thus £y x 
reduces itself to m x y v If we treat by S( )y the expression of p, 
we get by Syp 0 = 0 : 
Syp =. zSyy x ; 
and as Syy 2 = - m 1 by (12), we get 
Syp 
ffl, 
Hence 
(18) 
and 
(19) 
Syp 
p^po-yi— 
ip = ipo - ypSyp . 
We propose now to show, by a direct method, that ip 0 = 0, and 
for this purpose we apply the method of Hamilton, under the form 
by which it is explained in § 147 of Professor Tait’s Elementary 
Treatise on Quaternions . 
As (j>p for any vector p, and so also for p 0 , is perpendicular to y, 
we may represent <£p 0 by 
^p 0 = VyA, 
X being quite arbitrary, except that it must be different from y. 
This gives besides 
Sy4>p 0 = S ptfb'y = 0 , 
which is identically vanishing, the relation 
SX cj>p 0 = Sp 0 <£'X = 0 . 
We have also by hypothesis 
Sypo = 0 . 
From these two last equations we draw 
n being a scalar. 
npQ = V. y <£'X , 
2 Y 
VOL. XI. 
