(2015?浙江一模)如图,在直三棱柱A1B1C1-ABC中,AB⊥AC,AB=AC=2,AA1=4,点D是BC的中点.(1)求异面
(2015?浙江一模)如图,在直三棱柱A1B1C1-ABC中,AB⊥AC,AB=AC=2,AA1=4,点D是BC的中点.(1)求异面直线A1B与C1D所成角的余弦值;(2...
(2015?浙江一模)如图,在直三棱柱A1B1C1-ABC中,AB⊥AC,AB=AC=2,AA1=4,点D是BC的中点.(1)求异面直线A1B与C1D所成角的余弦值;(2)求平面ADC1与ABA1所成二面角的正弦值.
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(1)以{
,
,
}为单位正交基底建立空间直角坐标系A-xyz,
则由题意知A(0,0,0),B(2,0,0),C(0,2,0),
A1(0,0,4),D(1,1,0),C1(0,2,4),
∴
=(2,0,?4),
=(1,-1,-4),
∴cos<
,
>=
=
=
,
∴异面直线A1B与C1D所成角的余弦值为
.
(2)
=(0,2,0) 是平面ABA1的一个法向量,
设平面ADC1的法向量为
=(x,y,z),
∵
=(1,1,0),
=(0,2,4),
∴
AB |
AC |
AA1 |
则由题意知A(0,0,0),B(2,0,0),C(0,2,0),
A1(0,0,4),D(1,1,0),C1(0,2,4),
∴
A1B |
C1D |
∴cos<
A1B |
C1D |
| ||||
|
|
18 | ||||
|
3
| ||
10 |
∴异面直线A1B与C1D所成角的余弦值为
3
| ||
10 |
(2)
AC |
设平面ADC1的法向量为
m |
∵
AD |
AC1 |
∴
|