7在长方体ABCD-A1B1C1D1中,已知AA1=9,BC=6√3,N为BC的中点,则直线D1C1与平面A1B1N的距离是　　　　　.

答案:9

8已知正方体ABCD-A1B1C1D1的棱长为a,E,F分别是BB1,CD的中点,求点F到平面A1D1E的距离.

解:建立空间直角坐标系,如图所示,

　　则A1(a,0,a),D1(0,0,a),A(a,0,0),B(a,a,0),B1(a,a,a),E(a"," a"," a/2),F(0"," a/2 "," 0).

　　设平面A1D1E的法向量为n=(x,y, ),

　　则n·(A_1 D_1 ) ⃗=0,n·(A_1 E) ⃗=0,

　　即{■("(" x"," y"," z")·(-" a"," 0"," 0")" =0"," @"(" x"," y"," z")·" (0"," a",-" a/2)=0"," )┤

　　∴-ax=0,ay-a/2 =0.

　　∴{■(x=0"," @y=z/2 "," )┤令 =2,得n=(0,1,2).

　　又(FD_1 ) ⃗=(0",-" a/2 "," a),

∴所求距离d=("|" (FD_1 ) ⃗"·" n"|" )/("|" n"|" )=(3/2 a)/√5=(3√5)/10a.