WILSON AND LEWIS. — UEI^VIIVITY. 439 



For by (33) and (24), 



[(axb)xc]* = [ax(bxc)]*, 



(axb)'C* = a'(bxc')* = a'(b«c*). 



But since C is any 1 -vector, its complement C is any 2-vector. 

 Again, 



ax(b.C) = (axC)-b — (a-b)C. (36) 



For by (34), (22), and (30), 



[a.(bxc)]* = [(a-c) b]* ~ [(a-b) c]*, 



(_ 1)1(3-1) ap<(bxc)* = (_ 1)1(3-1) (axe*) -b _ (a.b)c*, 



ax(b«C) = (axC)«b — (a-b) C. 

 Again, 



(b.C)xA= -Cx(b.A). (37) 



For from (35), (30), and (24), 



[C-(axb)]* = [(b-C).a]*, 



(_ 1)2(3-2) cx(axb)* = (— l)i^3-i'(b-C)xa*, 



— Cx(bxa)* = — Cx(b-A)= (b-C)xA. 

 Again 



'& 



(b-C)-A = — b(C.A) + C.(bxA). (38) 



For from (36), (24), (32), (22), and (30), 



[(b.C)xaf - — [b-(Cxa)]* + [C(b.a)]*, 

 (b.C).A= - b(Cxa)*+ C.(b-a)*, 



= - b(C.A) + (-1)1(3-1) c.(bxA). 



These rules (33) to (38) involve every possible combination of three 

 vectors in three dimensional space. Since the formulas which we 

 ha\e used in deriving them, have the same form in Euclidean space, 

 the rules will be true in Euclidean space. The particular use of the 

 complement has implied a three dimensional space, and a similar use 

 of the complenent in a four dimensional space would obtain analogous 

 but different formulas; it should be observed, however, that the rules 

 here obtained (with the exception of (37)) must hold in space of four 

 dimensions, even when the three vectors in question do not lie wholly 

 in a three dimensional space. For consider (36) as a typical case. 

 Let b be a 1-vector which does not lie in the space of a and C; we 



