【答案】
分析:解法一(幾何法)(1)由已知可得BC⊥AB,又BC⊥PB由線面垂直的判定定理可得BC⊥平面PAB,則BC⊥PA.同理CD⊥PA,再中線面垂直的判定定理可得PA⊥平面ABCD.∠PCA為直線PC與底面ABCD所成的角,設(shè)M為AD中點(diǎn),連接EM,又E為PD中點(diǎn),可得EM∥PA,從而EM⊥底面ABCD.過(guò) M作AC的垂線MN,垂足為N,連接EN.則∠ENM為二面角E-AC-D的平面角,解Rt△EMN可得二面角E-AC-D的大小.
(2)若點(diǎn)E到平面PAF的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/0.png)
,則點(diǎn)D到平面PAF的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/1.png)
.過(guò) D作AF的垂線DG,垂足為G,可得DG為點(diǎn)D到平面PAF的距離設(shè)BF=x,由△ABF∽△DGA求出x的值,進(jìn)而得到結(jié)論.
解法二(向量法)(1)建立空間直角坐標(biāo)系A(chǔ)-xyz,求出平面AEC的一個(gè)法向量,和平面ACD的一個(gè)法向量,代入向量夾角公式,可得二面角E-AC-D的大�。�
(2)設(shè)F(2,t,0)(0≤t≤2),求出平面PAF的一個(gè)法向量,代入點(diǎn)E到平面PAF的距離d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/2.png)
,構(gòu)造方程求出t值,可得結(jié)論.
解答:解法一:(1)∵底面ABCD為正方形,
∴BC⊥AB,又BC⊥PB,AB∩PB=B
∴BC⊥平面PAB,
∴BC⊥PA.
同理可證CD⊥PA,由BC∩CD=C
∴PA⊥平面ABCD.
∴∠PCA為直線PC與底面ABCD所成的角,
∴PA=2 …(2分)
設(shè)M為AD中點(diǎn),連接EM,又E為PD中點(diǎn),可得EM∥PA,
從而EM⊥底面ABCD.過(guò) M作AC的垂線MN,垂足為N,連接EN.
由三垂線定理有EN⊥AC,∴∠ENM為二面角E-AC-D的平面角.…(4分)
在Rt△EMN中,可求得EM=1,MN=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/3.png)
∴tan∠ENM=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/5.png)
.
∴二面角E-AC-D的大小為arctan
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/6.png)
. …(6分)
(2)由E為PD中點(diǎn)可知,要使得點(diǎn)E到平面PAF的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/7.png)
,即要點(diǎn)D到平面PAF的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/8.png)
.
過(guò) D作AF的垂線DG,垂足為G,
∵PA⊥平面ABCD,∴平面PAF⊥平面ABCD,∴DG⊥平面PAF,
即DG為點(diǎn)D到平面PAF的距離.…9分
∴DG=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/9.png)
,∴AG=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/10.png)
. 設(shè)BF=x,由△ABF∽△DGA可得AB:BF=DG:GA,∴2:x=2:1,即x=1.∴在線段BC上存在點(diǎn)F,且F為BC中點(diǎn),使得點(diǎn)E到平面PAF的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/11.png)
. …(12分)
解法二:(1)∵底面ABCD為正方形,∴BC⊥AB,又BC⊥PB,∴BC⊥平面PAB,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/images12.png)
∴BC⊥PA.同理可證CD⊥PA,∴PA⊥平面ABCD.
∴∠PCA為直線PC與底面ABCD所成的角,
∴PA=2 …(2分)
建立如圖的空間直角坐標(biāo)系A(chǔ)-xyz,A(0,0,0),C(2,2,0),E(0,1,1),E(0,1,1)
設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/12.png)
=(x,y,z)為平面AEC的一個(gè)法向量,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/13.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/15.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/16.png)
.
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/17.png)
=(0,1,1),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/18.png)
=(2,2,0),
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/19.png)
令x=1則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/20.png)
=(1,-1,1)…(4分)
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/21.png)
=(0,0,2)是平面ACD的一個(gè)法向量,
設(shè)二面角E-AC-D的大小為 θ,
則cosθ=cos<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/23.png)
>=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/25.png)
.
∴二面角E-AC-D的大小為arccos
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/26.png)
.(6分)
(2)解:設(shè)F(2,t,0)(0≤t≤2),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/27.png)
=(a,b,c)為平面PAF的一個(gè)法向量,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/28.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/29.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/30.png)
⊥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/31.png)
.又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/32.png)
=(0,0,2),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/33.png)
=(2,t,0)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/34.png)
令a=t則b=-2,c=0得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/35.png)
=(t,-2,0). …(9分)
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/36.png)
=(0,1,1),∴點(diǎn)E到平面PAF的距離d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/38.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/39.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/40.png)
,解得t=1,即F(2,1,0),∴在線段BC上存在點(diǎn)F,使得點(diǎn)E到平面PAF的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125241399118466/SYS201310251252413991184019_DA/41.png)
,且F為BC中點(diǎn) …(12分)
點(diǎn)評(píng):本題考查的知識(shí)點(diǎn)是用空間向量求平面間的夾角,點(diǎn)到平面的距離,其中解法一的關(guān)鍵是熟練掌握空間線線垂直,線面垂直,面面垂直之間的相互轉(zhuǎn)化,解法二的關(guān)鍵是建立適當(dāng)?shù)目臻g坐標(biāo)系,將空間線面關(guān)系轉(zhuǎn)化為向量夾角問(wèn)題.