解 方法一 (1)建立如图所示的空间直角坐标系,
则A、B、C、D、P、E的坐标为A(0,0,0),B(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212826293.png)
,0,0)、C(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212826293.png)
,1,0)、D(0,1,0)、P(0,0,2)、
E(0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212842303.png)
,1),从而
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212857343.png)
=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212826293.png)
,1,0),
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212889345.png)
=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212826293.png)
,0,-2).
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212857343.png)
与
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212889345.png)
的夹角为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212951260.png)
,则cos
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212951260.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212982656.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212998452.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213123463.png)
,
∴AC与PB所成角的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213123463.png)
……………………………………7分
(2)由于N点在侧面PAB内,故可设N点坐标为(x,0,z),则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213154413.png)
=(-x,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212842303.png)
,1-z),由NE⊥平面PAC可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213185842.png)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232012132011436.png)
,化简得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213216776.png)
,∴
即N点的坐标为(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213247414.png)
,0,1),
从而N点到AB、AP的距离分别为1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213247414.png)
…………………14分
方法二 (1)设AC∩BD=O,
连接OE,AE,BD,则OE∥PB,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232012132942692.png)
∴∠EOA即为AC与PB所成的角或其补角.
在△AOE中,AO=1,OE=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212842303.png)
PB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213310403.png)
,AE=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212842303.png)
PD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213341412.png)
,
∴由余弦定理得cos∠EOA=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213357968.png)
,
即AC与PB所成角的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213123463.png)
.
(2)在平面ABCD内过D作AC的垂线交AB于F,则∠ADF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213388390.png)
.连接PF,则在Rt△ADF中,DF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213403636.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213403430.png)
,AF=AD·tan∠ADF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213419344.png)
.
设N为PF的中点,连接NE,则NE∥DF.
∵DF⊥AC,DF⊥PA,
∴DF⊥平面PAC,从而NE⊥平面PAC.
∴N点到AB的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212842303.png)
AP=1,N点到AP的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201212842303.png)
AF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823201213247414.png)
.