【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�,把異面直線轉(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)橹恍璁媯€(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),
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,C(a,a,0),D(0,2a,0),
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(1)
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,∴BE⊥PD
(2)由(1)知,
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=(-a,a,0)設(shè)
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所成角為θ則cosθ=
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(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
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∵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=
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a.
∴AE=
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=a.
∵PE=
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a.
∴ME=
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a.
連接AC
∵在△ACD中AD=2a,AC=
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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=
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.
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∴異面直線AE與CD所成角的余弦值為
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(9分)
(3)延長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∥
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AD,
∴GB=AB=a,∠PDA=30°,PA=
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a,AG=2a.
∴∠PGA=30°,
∴BF=
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=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,
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則A(0,0,0),B(a,0,0),
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,C(a,a,0),
D(0,2a,0),
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∴
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,
∴
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,
∴BE⊥PD(4分)
(2)由(1)知,
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=(-a,a,0)設(shè)
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所成角為θ
則cosθ=
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,
∴異面直線AE與CD所成角的余統(tǒng)值為
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.(9分)
(3)易知,CB⊥AB,CB⊥PA,
則CB⊥平面PAB.,∴
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是平面PAB的法向量.∴
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=(0,a,0).
又設(shè)平面PCD的一個(gè)法向量為
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,
則
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.而
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=(-a,a,0),
∴由
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=0.
得
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∴
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令y=1,,∴
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設(shè)向量
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所成角為θ,
則cosθ=
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.
∴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.