72 VECTOR ANALYSIS. 



Hence, since [<J> 2 ./3]x[<i?./3] and [$.ft]x/3 are parallel, 



Since a" 1 !^! is positive, we may set 



l^saa" 

 If we also set 



ft^-^.ft &=.p- 2 * 2 .& etc., 



/3_ i== p3>-i./3 } /3_ 2 = p*3>-2./3, etc., 



the vectors ft ft, ft, etc., /S.-p /3_ 2 , etc., will all lie in the plane 

 perpendicular to a, and we shall have 



ftxft = ftxft 



We may therefore set ft + ft = 2?ift . 



Multiplying by j)' 1 ^, and 



etc., 

 etc. 



Now, if ti > 1, and we lay off from a common origin the vectors 



ft ft, ft, etc., /?_!, /3_ 2 , etc., 



the broken line joining the termini of these vectors will be convex 

 toward the origin. All these vectors must therefore lie between two 

 limiting lines, which may be drawn from the origin, and which may 

 be described as having the directions of ft and /3_oo.* A vector 

 having either of these directions is unaffected in direction by 

 multiplication by <3?. In this case, therefore, $ is a tonic. If n < 1 

 we may obtain the same result by considering the vectors 



ft -ft, ft, -ft, ft, etc., -0-j, /3_ 2 , -0- 8 , etc, 

 except that a vector in the limiting directions will be reversed in 

 direction by multiplication by $, which implies that the two 

 corresponding coefficients of the tonic are negative. 



If 1 > n > l,t we may set 



n = cos q. 



Then 3 



Let us now determine y by the equation 



This gives /8. 1 = cosg/3 sin^y. 



Now a is one of the reciprocals of a, ft and y. Let ft and y' be the 



others. If we set 



we have 



* The termini of the vectors will in fact lie on a hyperbola, 

 t For the limiting cases, in which n=l, or n= - 1, see No. 156. 



