(1)证明略 (2) 当
=
时,A 1M⊥平面ADE
建立如图所示的空间直角坐标系D—xyz,
不妨设正方体的棱长为2,
则A(2,0,0),E(2,2,1),
F(0,1,0),A 1(2,0,2),D 1(0,0,2),
设平面AED的法向量为n 1=(x 1,y 1,z 1),
则n 1·
=(x 1,y 1,z 1)·(2,0,0)=0,
n 1·
=(x 1,y 1,z 1)·(2,2,1)=0,
∴2x 1=0,2x 1+2y 1+z 1=0.
令y 1=1,得n 1=(0,1,-2),
同理可得平面A 1FD 1的法向量n 2=(0,2,1).
∵n 1·n 2=0,∴n 1⊥n 2,
∴平面AED⊥平面A 1FD 1.
(2)解 由于点M在直线AE上,
设
=
=
(0,2,1)=(0,2
,
).
可得M(2,2
,
),∴
=(0,2
,
-2),
∵AD⊥A 1M,∴要使A 1M⊥平面ADE,
只需A 1M⊥AE,
∴
·
=(0,2
,
-2)·(0,2,1)=5
-2=0,
解得
=
.
故当
=
时,A 1M⊥平面ADE.