题目内容

如图,直四棱柱ABCD-A1B1C1D1中,平面A1BC⊥平面A1ABB1,AB=BC=2,AA1=2
2

(1)求证:BC⊥平面A1ABB1
(2)求直线A1B与平面A1AC成角的正弦值.
(1)∵直四棱柱ABCD-A1B1C1D1中,平面A1BC⊥平面A1ABB1
∴BC⊥AB,BC⊥BB1
又∵AB∩BB1=B,
∴BC⊥平面A1ABB1
(2)以DA为x轴,以DC为y轴,以DD1为z轴,建立空间直角坐标系,
∵直四棱柱ABCD-A1B1C1D1中,平面A1BC⊥平面A1ABB1,AB=BC=2,AA1=2
2

A1(2,0,2
2
)
,B(2,2,0),A(2,0,0),C(0,2,0),
AA1
=(0,0,2
2
),
AC
=(-2,2,0),
A1B
=(0,2,-2
2

设平面A1AC的法向量为
n
=(x,y,z),则
n
AA1
=0
n
AC
=0,
2
2
z=0
-2x+2y=0
,解得
n
=(1,1,0),
设直线A1B与平面A1AC成角为θ,
则sinθ=|cos<
n
A1B
>|=|
0+2+0
2
12
|=
6
6

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