to Non-Metrical Vector Algebra. 



133 



solved the inverse problem, viz. to bisect a given angle aOm. 

 In fact, if P be the pole of the chord am we have, by (2), 

 and replacing b by m, 



a + m 



P = 



1-f am* 



Multiply this equation sealarly by a or by m. The result 

 "will, in both cases, be 1. Thus Pa = Pm, and therefore, 



angle a, P = angle P, m. 



But P and S, the end-point of a + m, are collinear with (?. 

 Thus also 



aOS=SQm, 



giving the required bisection of the angle a Om . Remembering 

 'that S is the cross of aT m with niT a , this construction is 

 obtained at once, as shown in fig. 8. (Notice in passing- 

 Fig-. 8. 



that aS=Om, Oa=mS, and that all four arc unit vectors. 

 Thus OaSm is a generalized rhombus. In particular, if am= 0. 

 it will be a generalized square.) Thus far the bisection of 

 an angle. The solution of the inverse problem, originally 

 required, follows at once by remembering that 7\ the pole 

 of the required chord am, is collinear with 0, & Thus to 

 ^double aOb, draw the tangent at a, crossing the b-line in P : 



