380 HITCHCOCK. 



^i. That is, Foifii) coincides with ^i aside from a scalar factor. The 

 sufficiency of the conditions we may, if we prefer, show analytically, 

 by using components 641, 642, ^43, for ^a, etc. ; and remembering that 

 (456) does not vanish. 



From (12) and (9) we have 



(123)2 Fo{p) = ai(31p) (12p) + a2(12p) (23p) + a3(23p) (31p), (14) 



By writing respectively /34 and /Ss for p in this last equation, we obtain 

 the two equations of the first column of (13) in the form ^ 



(45ai) (314) (124) + (450.) (124) (234) + (45a3) (234) (314) = ., .^ 

 (45ai) (315) (125) + (45a2) (125) (235) + (45a3) (235) (315) = ^ ^ 



two linear homogeneous equations in the three quantities (45ai),. 

 (45a2), and (45a3). The two-row determinants from the matrix 

 of the coefficients cannot all be zero. For, identically, 



(I25) (235); (235) (sJs) I = (^34) (235) [(124) (315) - (125) (314)) 



= (234) (235) (415) (123) (16) 



whence, A\nth similar transformations ^ for the other two-row de- 

 terminants, and remembering that (123) does not vanish, we see that 

 the simultaneous vanishing of the three determinants would be equiva- 

 lent to 



(234) (235) (415) = 0, (314) (315) (425) = 0, (124) (125) (435) = 



(17) 



We cannot have (234) and (314) both zero, for ^s and ^i were taken 

 distinct in direction. We may, however, satisfy the first two condi- 

 tions by assuming (234) = (315) = 0. But (124) cannot vanish 

 with (234), (125) cannot vanish with (315), and (435) cannot vanish 

 with (315), since no four axes can lie in one plane. Similarly we may 

 exclude all other combinations of vanishing factors, one from each of 

 the three equations. 



5 Using products of vectors, we multiply both sides of (14) by the quaternion 

 0405 and take scalars. 



6 Using scalar products, the bracketed terms = S0i02{0iS030i0i — 0oS030i0i} 

 = S0i02V{V030iV0M = S0i020zS0i0504 = (123) (154). 



