题目内容
如图,直四棱柱ABCD-A1B1C1D1中,平面A1BC⊥平面A1ABB1,AB=BC=2,AA1=2
.
(1)求证:BC⊥平面A1ABB1;
(2)求直线A1B与平面A1AC成角的正弦值.
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
,
∴A1(2,0,2
),B(2,2,0),A(2,0,0),C(0,2,0),
∴
=(0,0,2
),
=(-2,2,0),
=(0,2,-2
)
设平面A1AC的法向量为
=(x,y,z),则
•
=0,
•
=0,
∴
,解得
=(1,1,0),
设直线A1B与平面A1AC成角为θ,
则sinθ=|cos<
,
>|=|
|=
.
∴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 |
∴
AA1 |
2 |
AC |
A1B |
2 |
设平面A1AC的法向量为
n |
n |
AA1 |
n |
AC |
∴
|
n |
设直线A1B与平面A1AC成角为θ,
则sinθ=|cos<
n |
A1B |
0+2+0 | ||||
|
| ||
6 |
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