解析:按照A,B两点在平面α的同侧和异侧分别讨论.

答案:C

4.如图,正方体ABCD-A1B1C1D1的棱长为1,O是平面A1B1C1D1的中心,则O到平面ABC1D1的距离是(　　)

A. 1/2 B.√2/4 C.√2/2 D.√3/2

解析:建立坐标系如图,则A(1,0,0),B(1,1,0),D1(0,0,1),O(1/2 "," 1/2 "," 1),

　　∴(AB) ⃗=(0,1,0),(AD_1 ) ⃗=(-1,0,1).

　　设n=(1,y,z)是平面ABC1D1的一个法向量,则

　　{■((AB) ⃗"·" n=y=0"," @(AD_1 ) ⃗"·" n="-" 1+z=0"," )┤解得y=0,z=1,

　　∴n=(1,0,1).又(OA) ⃗=(1/2 ",-" 1/2 ",-" 1),

　　∴点O到平面ABC1D1的距离为 ("|" (OA) ⃗"·" n"|" )/("|" n"|" )=(1/2)/√2=√2/4.

答案:B

★5.在长方体ABCD - A1B1C1D1中,底面是边长为2的正方形,高为4,则点A1到截面AB1D1的距离为(　　)

A. 8/3 B.3/8

C. 4/3 D.3/4

解析:利用V_(A" -" A_1 B_1 D_1 )=V_(A_1 "- " AB_1 D_1 ) 可求得点A1到截面AB1D1的距离为 4/3.

答案:C