(1)证明FD
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025505183.png)
平面PAF即可.
(2)取AD的四分之一分点N,使
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025521674.png)
m则EN//DF,然后再取PA的四分之一分点,使
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025552648.png)
,即是所求G点位置.易证EG//平面PFD.
(3)利用空间向量法求解即可.要把二面角两个面的法向量求出来,然后再求法向量的夹角.
解:(1)证明:连接AF,则AF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025568344.png)
,DF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025568344.png)
,
又AD=2,∴DF
2+AF
2=AD
2,∴DF⊥AF.又PA⊥平面ABCD,
∴DF⊥PA,又PA∩AF=A,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232250257081399.png)
……………4分
(2)过点E作EH∥FD交AD于点H,则EH∥平面PFD且AH=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025458303.png)
AD.
再过点H作HG∥DP交PA于点G,则HG∥平面PFD且AG=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025458303.png)
AP,
∴平面EHG∥平面PFD.∴EG∥平面PFD.
从而满足AG=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025458303.png)
AP的点G为所求.………………8分
(3)建立如图所示的空间直角坐标系,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232250258803173.jpg)
因为PA⊥平面ABCD ,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025895496.png)
是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025380365.png)
与平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025396526.png)
所成的角.又有已知得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025973656.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026004528.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232250260201652.png)
.
设平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026036447.png)
的法向量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026067688.png)
,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026082914.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026098824.png)
,令
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026129289.png)
,解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026145570.png)
.
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026160853.png)
.又因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026176637.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026207412.png)
是平面
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026223439.png)
的法向量,易得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225026238685.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232250262701912.png)
.
由图知,所求二面角
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025443551.png)
的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823225025490421.png)
.……………………12分