题目内容
(本小题满分12分)
如图,已知三棱锥P—ABC中,PA⊥平面ABC,
AB⊥AC,PA=AC=
AB,N为AB
上一点,
AB=4AN,M,S分别为PB,BC的中点.
(I)证明:CM⊥SN;
(II)求SN与平面CMN所成角的大小.
如图,已知三棱锥P—ABC中,PA⊥平面ABC,
AB⊥AC,PA=AC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639105213.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163912172.gif)
AB=4AN,M,S分别为PB,BC的中点.
(I)证明:CM⊥SN;
(II)求SN与平面CMN所成角的大小.
(I)略
(II)SN与平面CMN所成角为45°.
(II)SN与平面CMN所成角为45°.
设PA=1,以A为原点,射线AB,AC,AP分别为x,y,z轴正向建立空间直角坐标系,如图.
则P(0,0,1),C(0,1,0),B(2,0,0),M(1,0,
),
N(
,0,0),S(1,
,
0),(I)证明:
CM=(1,-1,
),
SN=(-
,-
,0),
因为
CM·
SN=-
+
+0=0,所以CM⊥SN.
(II)解:
NC=(-
,1,0),
设a=(x,y,z)为平面CMN的一个法向量,则![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639947635.gif)
令x=2,得a=(2,1,-2),因为|cos(a,
SN)|=|
|=
,
所以SN与平面CMN所成角为45°.
则P(0,0,1),C(0,1,0),B(2,0,0),M(1,0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163940165.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163944889.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163946490.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163966790.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163974591.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
(II)解:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163982388.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639136206.gif)
设a=(x,y,z)为平面CMN的一个法向量,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639947635.gif)
令x=2,得a=(2,1,-2),因为|cos(a,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082315163946490.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151639979358.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823151640010238.gif)
所以SN与平面CMN所成角为45°.
![](http://thumb2018.1010pic.com/images/loading.gif)
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