88 VECTOR ANALYSIS. 



6. To reduce a given bivector r to the above form, we may set 



r.r=(cos(?-H sin g) 2 [a + //3].[a + */3] 

 = (cos 2q + 1 sin 2q) (a. a ft. ft) 



where a and 6 are scalars, which we may regard as known. The 

 value of q may be determined by the equation 



tan 2(7 = -, 



a 



the quadrant to which 2g belongs being determined so as to give 

 sin 2q and cos 2q the same signs as b and a. Then a and ft will be 

 given by the equation 



a + i/3 (cos q i sin g)t. 



The solution is indeterminate when the real and imaginary parts of 

 the given bivector are perpendicular and equal in magnitude. In this 

 case the directional ellipse is a circle, and the bivector may be called 

 circular. The criterion of a circular bivector is 



r.r = 0. 



It is especially to be noticed that from this equation we cannot 

 conclude that 



1=0, 



as in the analysis of real vectors. This may also be shown by ex- 

 pressing t in the form xi + yj+zk, in which x, y, z are biscalars. The 

 equation then becomes 



which evidently does not require x, y, and z to vanish, as would be 

 the case if only real values are considered. 



7. Def. We call two vectors p and <r perpendicular when p.0- = 0. 

 Following the same analogy, we shall call two bivectors t and $ 

 perpendicular, when 



In considering the geometrical signification of this equation, we shall 

 first suppose that the real and imaginary components of t and $ lie in 

 the same plane, and that both t and 3 have not real directions. It is 

 then evidently possible to express them in the form 



where m and m' are biscalar, a and ft are at right angles, and a 

 parallel with ft. Then the equation r.$ = requires that 



/3.p = 0, and a.p+p.a=0. 



This shows that the directional ellipses of the two bivectors are 

 similar and the angular direction from the real to the imaginary 



