例2证明向量c与向量(a·c)b-(b·c)a垂直 证I(a·c)b-(b·cd (a·c)b·c-(b·c)d =(C·b)4·c-l·c =0 (a·c)b-(b·)d⊥c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        [(  ) − (  ) ]⊥