试题分析:(1)取BD中点Q,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240123574621144.png)
三点共线,即Q与O重合。
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240123574311016.png)
面PBD
(2)因为AC
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357493183.png)
面PBD,而
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357509439.png)
面ABCD,所以面ABCD
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357493183.png)
面PBD,则P点在面ABCD上的射影点在交线BD上(即在射线OD上),所以PO与平面ABCD所成的角
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357540799.png)
。以O为坐标原点,OA为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357555266.png)
轴,OB为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357571310.png)
轴建空间直角坐标系。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240123575871605.png)
,因为AC
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357493183.png)
面PBD,所以面PBD的法向量
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357618772.png)
,设面PAB的法向量
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357633717.png)
,又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357649875.png)
,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357665595.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357680618.png)
①,又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240123576961152.png)
,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357711562.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357727825.png)
②, 在①②中令
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357743476.png)
,可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357758436.png)
,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357774811.png)
所以二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824012357415533.png)
的余弦值
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240123574461269.png)
点评:典型题,立体几何题,是高考必考内容,往往涉及垂直关系、平行关系、角、距离、体积的计算。在计算问题中,有“几何法”和“向量法”。利用几何法,要遵循“一作、二证、三计算”的步骤,将立体问题转化成平面问题,是解决立体几何问题的一个基本思路。通过就落实党的坐标系,利用空间向量,免去了繁琐的逻辑推理过程,对计算能力要求较高。