解决立体几何问题的,主要有两个策略,一是不建立坐标系,直接利用空间向量基本定理,即将有关向量用空间一组基底表示出来,然后通过向量的有关运算求解;二是建立空间坐标系,通过向量的坐标运算解决问题
方法一:
(I)证明:∵平面PCD⊥平面ABCD,又∵平面PCD∩平面ABCD=CD,BC在平面ABCD内 ,BC⊥CD,∴BC⊥平面PCD.∴PD⊥BC. …………6分
(II)解:取PD的中点E,连接CE、BE,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232212303826954.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230397549.png)
为正三角形,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230428605.png)
由(I)知BC⊥平面PCD,∴CE是BE在平面PCD内的射影,∴BE⊥PD.
∴∠CEB为二面角B—PD—C的平面角. …………9分
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232212304441068.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230460841.png)
…………12分
方法二:(I)证明:取CD的中点为O,连接PO,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232212304917336.png)
∵PD=PC,∴PO⊥CD,∵平面PCD⊥平面ABCD,
平面PCD∩平面ABCD=CD,∴PO⊥平面ABCD,如图,在平面ABCD内,过O作OM⊥CD交AB于M,以O为原点,OM、OC、OP分别为x、y、z轴,建立空间直角坐标系O—xyz,
由B(2,1,0),C(0,1,0),D(0,-1,0),
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230506613.png)
…………4分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232212305381335.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230553582.png)
…6分
(II)解:取PD的中点E,连接CE、BE,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230569827.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230600554.png)
为正三角形,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232212306161637.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221230647547.png)
为二面角B—PD—C的平面角.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232212306781193.png)