解法一: (I)由已知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214109287750.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214109431139.gif)
∴PG=4
如图所示,以G点为原点建立空间直角坐标系
o—
xyz,则
B(2,0,0),C(0,2,0),P(0,0,4)
故E(1,1,0)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121410959750.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214109751383.gif)
∴异面直线GE与PC所成的角为arccos
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121410990333.gif)
(II)平面PBG的单位法向量
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411006528.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214110211644.gif)
∴点D到平面PBG的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411037581.gif)
(III)设F(0,y , z)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214110683079.gif)
在平面PGC内过F点作FM⊥GC,M为垂足,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411068653.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411099687.gif)
解法二:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214111317380.jpg)
(I)由已知
∴PG=4
在平面ABCD内,过C点作CH//EG交AD于H,连结PH,则∠PCH(或其补角)就是异面直线GE与PC所成的角.
在△PCH中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411162672.gif)
由余弦定理得,cos∠PCH=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121410990333.gif)
,∴异面直线GE与PC所成的角为arccos
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121410990333.gif)
(II)∵PG⊥平面ABCD,PG
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411224135.gif)
平面PBG ∴平面PBG⊥平面ABCD
在平面ABCD内,过D作DK⊥BG,交BG延长线于K,
则DK⊥平面PBG ∴DK的长就是点D到平面PBG的距离
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411240989.gif)
在△DKG,DK=DGsin45°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411255223.gif)
∴点D到平面PBG的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823121411255223.gif)
(III)在平面ABCD内,过D作DM⊥GC,M为垂足,连结MF,又因为DF⊥GC
∴GC⊥平面MFD,∴GC⊥FM
由平面PGC⊥平面ABCD,∴FM⊥平面ABCD ∴FM//PG
由GM⊥MD得:GM=GD·cos45°=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408231214113331446.gif)