G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=9. 63 
¢ P= —A(1+2+2’) 
(50 ter) 
Q’ =hx(l +2). 
Further we know that %,4=0 is the resultant of the equation (o—h)p=0 
written for three directions X, p, v, for which this equation is satisfied, or 
VABA=0, Vuop=0, Vrav=0 
The unit vectors A, p, v, being so defined, we know that they are treble 
rectangular, and we may suppose h,, h, hs corresponding respectively to 
h, p, v. Then we have: 
BA=hvV, GBu=hp, B=hy, 
and from this, supposing 
€ =— DSe , 
we deduce 
‘= = Zh She’. 
And as E=—Seae’, we get 
(51) E= 2h Sr’ . 
This expression, owing to 2S’\c =1, may be put under the form : 
(51 bis) >(h, —E)Se’ =0, 
and it shows that the value of E must not be greater than h, nor smaller than 
h;, lest all the three terms would be of the same sign. 
Therefore we must always have 
hg SBD S hy. 
We have now prepared all that is required for the.discussion oF the three 
conditions (a), (6), (c). As to (c) we may write 
FE = (h,— E)(h,—E)(E=h,) , 
and considering the limits of E just above we see that the product (h,—E) 
(E—A,) will always be positive, and the sign of 7E will papend on that of 
(h,—E). Thus (c) becomes 
(0) — (HPE-Q)(y-E) > 0. 
Putting for —P’, Q’ their expression in z, and suppressing the factor h?, 
— we get . eo WE O40 cs 
[(l+2+2)E—zh,1+2)][hez—E] > 0, 
oras 1+2+2°>0: 
() [ Enz it)] [ye—E] > 0. 
