【答案】
分析:解法一:
(1)欲證直線與直線垂直,可用先證直線與平面垂直.∵BA⊥AD,BA⊥PA,∴BA⊥平面PAD.∴PD⊥BA.又∵PD⊥AE,∴PD⊥平面BAE,∴PD⊥BE.
(2)求異面直線所成的角,可以做適當(dāng)?shù)钠揭疲旬惷嬷本€轉(zhuǎn)化為相交直線,然后在相關(guān)的三角形中借助正弦或余弦定理解出所求的角.平移時(shí)主要是根據(jù)中位線和中點(diǎn)條件,或者是特殊的四邊形,三角形等.過(guò)點(diǎn)E作EM∥CD交PC于M,連接AM,則AE與ME所成角即為AE與CD所成角.
(3)二面角的度量關(guān)鍵在于找出它的平面角,構(gòu)造平面角常用的方法就是三垂線法.延長(zhǎng)AB與DC相交于G點(diǎn),連PG,則面PAB與面PCD的交線為PG,易知CB⊥平面PAB,過(guò)B作BF⊥PG于F點(diǎn),連CF,則CF⊥PG,∴∠CFB為二面角C-PG-A的平面角
解法二:
在含有直線與平面垂直垂直的條件的棱柱、棱錐、棱臺(tái)中,也可以建立空間直角坐標(biāo)系,設(shè)定參量求解.這種解法的好處就是:1、解題過(guò)程中較少用到空間幾何中判定線線、面面、線面相對(duì)位置的有關(guān)定理,因?yàn)檫@些可以用向量方法來(lái)解決.2、即使立體感稍差一些的學(xué)生也可以順利解出,因?yàn)橹恍璁?huà)個(gè)草圖以建立坐標(biāo)系和觀察有關(guān)點(diǎn)的位置即可.則A(0,0,0),B(a,0,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/0.png)
,C(a,a,0),D(0,2a,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/1.png)
(1)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/2.png)
,∴BE⊥PD
(2)由(1)知,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/3.png)
=(-a,a,0)設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/4.png)
所成角為θ則cosθ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/5.png)
(3)利用平面PAB與平面PCD的法向量所成的角,去求平面PAB與平面PCD所成的銳二面角的正切值.
解答:解法一:(1)∵∠BAD=90°,∴BA⊥AD
∵PA⊥底面ABCD,BA⊥PA.又∵PA∩AD=A,BA⊥PA.又∵PA∩AD=A,
∴BA⊥平面PAD.
∵PD?平面PAD.
∴PD⊥BA.又∵PD⊥AE,且BA∩AE=A,
∴PD⊥平面BAE
∴PD⊥BE,即BE⊥PD.(4分)
(2)過(guò)點(diǎn)E作EM∥CD交PC于M,連接AM,則AE與ME所成角即為AE與CD所成角
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/images6.png)
∵PA⊥底面ABCD,且PD與底面ABCD成30°角.
∴∠PDA=30°.
∴在Rt△PAD中,∠PAD=90°,∠PDA=30°,AD=2a
∴PA=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/6.png)
a.
∴AE=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/7.png)
=a.
∵PE=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/8.png)
a.
∴ME=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/9.png)
a.
連接AC
∵在△ACD中AD=2a,AC=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/10.png)
a,
AD
2=AC
2+CD
2∴∠ACD=90°,∴CD⊥AC,∴ME⊥AC
又∵PA⊥底面ABCD,
∴PA⊥CD,∴ME⊥PA.
∴ME⊥平面PAC.∵M(jìn)A?平面PAC,
∵M(jìn)E⊥AM.
∴在Rt△AME中,cos∠MEA=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/11.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/images13.png)
∴異面直線AE與CD所成角的余弦值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/12.png)
(9分)
(3)延長(zhǎng)AB與DC相交于G點(diǎn),連PG,則面PAB
與面PCD的交線為PG,易知CB⊥平面PAB,過(guò)B作BF⊥PG于F點(diǎn),連CF,則CF⊥PG,
∴∠CFB為二面角C-PG-A的平面角,
∵CB∥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/13.png)
AD,
∴GB=AB=a,∠PDA=30°,PA=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/14.png)
a,AG=2a.
∴∠PGA=30°,
∴BF=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/15.png)
=2,
∴平面PAB與平面PCD所成的二面角的正切值為2.(14分)
解法二:(1)如圖建立空間直角坐標(biāo)系,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/images18.png)
則A(0,0,0),B(a,0,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/16.png)
,C(a,a,0),
D(0,2a,0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/17.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/18.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/19.png)
,
∴BE⊥PD(4分)
(2)由(1)知,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/20.png)
=(-a,a,0)設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/21.png)
所成角為θ
則cosθ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/22.png)
,
∴異面直線AE與CD所成角的余統(tǒng)值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/23.png)
.(9分)
(3)易知,CB⊥AB,CB⊥PA,
則CB⊥平面PAB.,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/24.png)
是平面PAB的法向量.∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/25.png)
=(0,a,0).
又設(shè)平面PCD的一個(gè)法向量為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/26.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/27.png)
.而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/28.png)
=(-a,a,0),
∴由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/29.png)
=0.
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/30.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/31.png)
令y=1,,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/32.png)
設(shè)向量
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/33.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/34.png)
所成角為θ,
則cosθ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180637057934832/SYS201310241806370579348017_DA/35.png)
.
∴tanθ=2.
∴平面PAB與平面PCD所成銳二面角的正切值為2.(14分)
點(diǎn)評(píng):本小題主要考查空間線面關(guān)系、面面關(guān)系、二面角的度量等知識(shí),考查數(shù)形結(jié)合、化歸與轉(zhuǎn)化的數(shù)學(xué)思想方法,以及空間想象能力、推理論證能力和運(yùn)算求解能力.