因为(MN) ⃗·(A"'" C) ⃗=0,

　　所以MN⊥A'C.又(MN) ⃗·(BB) ⃗'=0,所以MN⊥BB'.

★10.在棱长为1的正方体ABCD - A1B1C1D1中,E,F分别为棱AB和BC的中点,试在棱B1B上找一点M,使得D1M⊥平面EFB1.

解:以D为坐标原点,(DA) ⃗,(DC) ⃗,(DD_1 ) ⃗的方向分别为x轴,y轴,z轴的正方向建立如图所示的空间直角坐标系Dxyz.

　　有A(1,0,0),B1(1,1,1),C(0,1,0),D1(0,0,1),E(1"," 1/2 "," 0),F(1/2 "," 1"," 0),设点M(1,1,m),

　　则(EF) ⃗=("-" 1/2 "," 1/2 "," 0),(B_1 E) ⃗=(0",-" 1/2 ",-" 1),(D_1 M) ⃗=(1,1,m-1).

　　∵D1M⊥平面EFB1,

　　∴(D_1 M) ⃗⊥(EF) ⃗,且(B_1 E) ⃗⊥(D_1 M) ⃗,

　　∴(D_1 M) ⃗·(EF) ⃗=0,(D_1 M) ⃗·(B_1 E) ⃗=0,

　　∴{■("-" 1/2+1/2+"(" m"-" 1")·" 0=0"," @1×0"-" 1/2+1"-" m=0"," )┤

　　∴m=1/2.

　　故取B1B的中点M,能满足D1M⊥平面EFB1.