第2课时　直线到平面的距离、平面到平面的距离

1.已知,在长方体ABCD-A1B1C1D1中,棱A1A=5,AB=12,则直线B1C1到平面A1BCD1的距离是(　　)

A.5 B. 13/2 C.60/13 D.8

解析:如图所示,建立空间直角坐标系,设AD=a,

则A1(a,0,5),B(a,12,0),C(0,12,0),D1(0,0,5),B1(a,12,5),C1(0,12,5),

∴(A_1 B) ⃗=(0,12,-5),(BC) ⃗=(-a,0,0),(B_1 B) ⃗=(0,0,-5).

设平面A1BCD1的法向量为n=(x,y,z),

则{■(n"·" (A_1 B) ⃗=0"," @n"·" (BC) ⃗=0"," )┤

即{■(12y"-" 5z=0"," @ax=0"." )┤

令y=5,则n=(0,5,12).

因此直线B1C1到平面A1BCD1的距离为

d=("|" (B_1 B) ⃗"·" n"|" )/("|" n"|" )=60/13.

答案:C

2.在棱长为1的正方体ABCD-A1B1C1D1中,M,N分别是线段BB1,B1C1的中点,则直线MN和平面ACD1的距离是(　　)

A. 1/2 B.√2/2

C. 1/3 D.√3/2

解析:如图所示,建立空间直角坐标系,则A(1,0,0),D1(0,0,1),M(1"," 1"," 1/2),N(1/2 "," 1"," 1),C(0,1,0).