法一:由已知得B1D⊥平面AC1, 又CF⊂平面AC1,∴B1D⊥CF, 故若CF⊥平面B1DF,则必有CF⊥DF. 设AF=x(0<x<3a),则CF2=x2+4a2, DF2=a2+(3a-x)2,又CD2=a2+9a2=10a2, ∴10a2=x2+4a2+a2+(3a-x)2, 解得x=a或2a. 法二:分别以BA、BC、BB1所在直线为x轴、y轴、z轴建立空间直角坐标系B-xyz, 则B(0,0,0),B1(0,0,3a),设F(a,0,m),D,C(0,a,0), =(a,-a,m),=,=(a,0,m-3a), ∵CF⊥面B1DF,∴CF⊥B1F,⊥,即·=0,·=0, 可得2a2+m(m-3a)=0,解得m=a或2a. |