题目内容
在四棱锥P—ABCD中,底面ABCD是正方形,侧棱PD⊥平面ABCD,AB=PD=a.点E为侧棱PC的中点,又作DF⊥PB交PB于点F.则PB与平面EFD所成角为( )
A.30° | B.45° | C.60° | D.90° |
D
建立空间直角坐标系D—xyz,D为坐标原点.P(0,0,a),B(a,a,0),
=(a,a,-a),又
=
,
=0+
-
=0,
所以PB⊥DE.由已知DF⊥PB,又DF∩DE=D,
所以PB⊥平面EFD,所以PB与平面EFD所成角为90°,选D.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824045208903397.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824045208919407.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824045208934744.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824045208950503.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824045208965406.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824045208965406.png)
所以PB⊥DE.由已知DF⊥PB,又DF∩DE=D,
所以PB⊥平面EFD,所以PB与平面EFD所成角为90°,选D.
![](http://thumb.zyjl.cn/images/loading.gif)
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