【答案】
分析:法一、(1)要證明線面平行,關(guān)鍵是在平面內(nèi)找到一條可能與已知直線平行的直線,觀察到平面ADD
1A
1中三條已知直線與PC都不平行,故我們要考慮在平面ADD
1A
1中做一條與PC可能平行直線輔助線,然后再進(jìn)行證明.
(2)要求二面角的余弦,要先構(gòu)造出二面角的平面角,然后利用解三角形的方法,求出這個(gè)平面角的余弦值,進(jìn)而給出二面角的余弦值.
(3)要求三棱錐的體積,只要求出底面的面積,及對(duì)應(yīng)的高代入棱錐體積公式,即可求解.
法二、構(gòu)造空間直角坐標(biāo)系,求出各點(diǎn)的坐標(biāo),進(jìn)行求出相應(yīng)直線的方向向量和平面的法向量,利用向量法進(jìn)行求解.
解答:![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/images0.png)
解:法一:(Ⅰ)證明:取CD的中點(diǎn)K,連接MK,NK
∵M(jìn),N,K分別為AK,CD
1,CD的中點(diǎn)
∵M(jìn)K∥AD,NK∥DD
1∴MK∥面ADD
1A
1,NK∥面ADD
1A
1,又MK與NK交于K
∴面MNK∥面ADD
1A
1,
∴MN∥面ADD
1A
1(Ⅱ)設(shè)F為AD的中點(diǎn)
∵P為A
1D
1的中點(diǎn)∴PF∥D
1D∴PF⊥面ABCD
作FH⊥AE,交AE于H,連接PH,則由三垂線定理得AE⊥PH
從而∠PHF為二面角P-AE-D的平面角.
在Rt△AEF中,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/0.png)
,
從而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/1.png)
在Rt△PFH中,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/2.png)
故:二面角P-AE-D的大小為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/3.png)
(Ⅲ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/4.png)
作DQ⊥CD
1,交CD
1于Q,由A
1D
1⊥面CDD
1C
1得A
1C
1⊥DQ
∴DQ⊥面BCD
1A
1∴在Rt△CDD
1中,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/5.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/8.png)
方法二:以D為原點(diǎn),DA,DC,DD
1所在直線分別為x軸,y軸,z軸,建立直角坐標(biāo)系,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/images10.png)
則A(a,0,0),B(a,2a,0),C(0,2a,0),A
1(a,0,a),D
1(0,0,a)
∵E,P,M,N分別是BC,A
1D
1,AE,CD
1的中點(diǎn)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/9.png)
,
(Ⅰ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/10.png)
取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/11.png)
,顯然
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/12.png)
面ADD
1A
1![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/13.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/14.png)
又MN∉面ADD
1A
1∴MN∥面ADD
1A
1(Ⅱ)過P作PH⊥AE,交AE于H,取AD的中點(diǎn)F,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/15.png)
∵設(shè)H(x,y,0),則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/16.png)
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/17.png)
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/18.png)
,及H在直線AE上,可得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/19.png)
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/20.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/21.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/22.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/23.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/24.png)
與
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/25.png)
所夾的角等于二面角P-AE-D的大小
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/26.png)
故:二面角P-AE-D的大小為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/27.png)
(Ⅲ)設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/28.png)
為平面DEN的法向量,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/29.png)
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/30.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/31.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/32.png)
∴可取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/33.png)
∴P點(diǎn)到平面DEN的距離為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/34.png)
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/35.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/36.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212856241342422/SYS201310232128562413424021_DA/37.png)
∴
點(diǎn)評(píng):判斷或證明線面平行的常用方法有:①利用線面平行的定義(無公共點(diǎn));②利用線面平行的判定定理(a?α,b?α,a∥b⇒a∥α);③利用面面平行的性質(zhì)定理(α∥β,a?α⇒a∥β);④利用面面平行的性質(zhì)(α∥β,a?α,a?,a∥α⇒?a∥β).
求二面角,關(guān)鍵是構(gòu)造出二面角的平面角,常用的方法有利用三垂線定理和通過求法向量的夾角,然后再將其轉(zhuǎn)化為二面角的平面角.本題也可以用空間向量來解決,其步驟是:建立空間直角坐標(biāo)系⇒明確相關(guān)點(diǎn)的坐標(biāo)⇒明確相關(guān)向量的坐標(biāo)⇒通過空間向量的坐標(biāo)運(yùn)算求解.