设平面A1B1N的法向量为n=(x,y, ),

　　则{■(n"·" (A_1 B_1 ) ⃗=0"," @n"·" (B_1 N) ⃗=0"," )┤即{■(ay=0"," @"-" 3√3 x"-" 9z=0"," )┤

　　取x=3,则y=0, =-√3,

　　所以平面A1B1N的一个法向量为n=(3,0,-√3),

　　所以点D1到平面A1B1N的距离d=("|" (D_1 A_1 ) ⃗"·" n"|" )/("|" n"|" )=(18√3)/(2√3)=9.

　　又因为D1C1∥平面A1B1N,

　　所以直线D1C1到平面A1B1N的距离仍是9.

　　【答案】9

6.在正三棱柱ABC-A1B1C1中,所有棱长均为1,则点B1到平面ABC1的距离d为　　　　.

　　【解析】

　　(法一)建立如图所示的空间直角坐标系,则点C(0,0,0),A(√3/2 "," 1/2 "," 0),B(0,1,0),B1(0,1,1),C1(0,0,1),

　　∴(C_1 A) ⃗=(√3/2 "," 1/2 ",-" 1),(C_1 B_1 ) ⃗=(0,1,0),(C_1 B) ⃗=(0,1,-1),

　　设平面ABC1的一个法向量为n=(x,y,1),

　　则{■((C_1 A) ⃗"·" n=0"," @(C_1 B) ⃗"·" n=0"," )┤即{■(√3/2 x+1/2 y"-" 1=0"," @y"-" 1=0"," )┤

　　解得n=(√3/3 "," 1"," 1),

　　∴d=("|" (C_1 B_1 ) ⃗"·" n"|" )/("|" n"|" )=1/√(1/3+1+1)=√21/7.

　　(法二)V_(B_1 "-" ABC_1 )=V_(A"-" BB_1 C_1 ),

　　V_(A"-" BB_1 C_1 )=1/3 S_("△" BB_1 C_1 )×√3/2AB=√3/12,

　　∵V_(B_1 "-" ABC_1 )=1/3 S_("△" ABC_1 )·d,S_("△" ABC_1 )=1/2AB·√(1^2+1^2 "-" (1/2)^2 )=√7/4,∴d=√21/7.

　　【答案】√21/7

7.如图所示的多面体是由底面为ABCD的长方体被截面AEC1F所截得到的,其中AB=4,BC=2,CC1=3,BE=1,四边形AEC1F为平行四边形.

(1)求BF的长;

(2)求点C到平面AEC1F的距离.

【解析】