例2证明向量与向量(a.c)b-(b·c)d垂直 证I(a·c)b-(b·c)l =I(a·c)bc-(b·c)a·dl =(c·b川a·c-a·dl =0 I(a·c)b-(b·c)ilc 上页例 2 证明向量c 与向量 a c b b c a ( ) − ( ) 垂直. 证 a c b b c a c [( ) −( ) ] [(a c)b c (b c)a c] = − (c b)[a c a c] = − = 0 a c b b c a c [( ) −( ) ]⊥