解析:分别以PA,PB,PC所在的直线为x轴,y轴, 轴建立空间直角坐标系(图略),则A(1,0,0),B(0,1,0),C(0,0,1).可以求得平面ABC的一个法向量为n=(1,1,1),则d=("|" (PA) ⃗"·" n"|" )/("|" n"|" )=√3/3.

答案:D

4在正三棱柱ABC-A1B1C1中,若AB=AA1=4,点D是AA1的中点,则点A1到平面DBC1的距离是(　　)

A.√2 B.√2/2 C.√2/3 D.√2/4

答案:A

5已知直线l过原点,一个方向向量为n=(1,1,1),则点A(0,0,3)到直线l的距离为　　　　　.

答案:√3

6已知A(2,3,1),B(4,1,2),C(6,3,7),D(-5,-4,8),则点D到平面ABC的距离为　　　　　.

解析:设平面ABC的法向量n=(x,y, ),

　　则{■(n"·" (AB) ⃗=0"," @n"·" (AC) ⃗=0"," )┤即{■("(" x"," y"," z")·(" 2",-" 2"," 1")" =0"," @"(" x"," y"," z")·(" 4"," 0"," 6")" =0"." )┤

　　∴可取n=("-" 3/2 ",-" 1"," 1).又(AD) ⃗=(-7,-7,7),

　　∴点D到平面ABC的距离d=("|" (AD) ⃗"·" n"|" )/("|" n"|" )=(49√17)/17.

答案:(49√17)/17