【解析】

　　(1)建立如图所示的空间直角坐标系,则点D(0,0,0),B(2,4,0),A(2,0,0),C(0,4,0),E(2,4,1),C1(0,4,3),设点F(0,0, ).

　　∵四边形AEC1F为平行四边形,

　　∴由(AF) ⃗=(EC_1 ) ⃗得(-2,0, )=(-2,0,2),

　　∴ =2,∴点F(0,0,2),∴(BF) ⃗=(-2,-4,2),|(BF) ⃗|=2√6,即BF的长为2√6.

　　(2)设n1为平面AEC1F的法向量,

　　显然n1不垂直于平面ADF,

　　故可设n1=(x,y,1),∴{■(n_1 "·" (AE) ⃗=0"," @n_1 "·" (AF) ⃗=0"," )┤

　　即{■(4y+1=0"," @"-" 2x+2=0"," )┤解得{■(x=1"," @y="-" 1/4 "," )┤∴n1=(1",-" 1/4 "," 1).

　　又∵(CC_1 ) ⃗=(0,0,3),

　　∴点C到平面AEC1F的距离为d=("|" (CC_1 ) ⃗"·" n_1 "|" )/("|" n_1 "|" )=(4√33)/11.

拓展提升(水平二)

8.如图,A是正方形BCDE外一点,AE⊥平面BCDE,且AE=CD=a,G,H分别是BE,ED的中点,则GH到平面ABD的距离是(　　).

　　A.√3/3a B.√3/6a

　　C.√2/3a D.√2/6a

　　【解析】如图,由题意知GH∥平面ABD,以E为坐标原点,EB,ED,EA所在直线分别为x轴,y轴, 轴建立空间直角坐标系,则点E(0,0,0),G(a/2 "," 0"," 0),H(0"," a/2 "," 0),A(0,0,a),B(a,0,0),D(0,a,0),(AB) ⃗=(a,0,-a),(AD) ⃗=(0,a,-a).设平面ABD的一个法向量为n=(x,y, ),

　　则{■(n"·" (AB) ⃗=ax"-" az=0"," @n"·" (AD) ⃗=ay"-" az=0"," )┤

　　所以x=y= .

　　可取n=(1,1,1),

又因为(GB) ⃗=(a/2 "," 0"," 0),