例2证明向量乙与向量(a·d)b-(b·)d垂直 证(a·C)b-(b-)l乙 =[(a·c)b·-(b·c)·dl (·b)la·c-a·l 0 (a·c)b-(b·d)dc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        [(  ) − (  ) ]⊥