試題分析:解法一(向量法)
(I)建立如圖所示的空間直角坐標系A(chǔ)-xyz,分別求出直線PF與FD的平行向量,然后根據(jù)兩個向量的數(shù)量積為0,得到PF⊥FD;
(2)求出平面PFD的法向量(含參數(shù)t),及EG的方向向量,進而根據(jù)線面平行,則兩個垂直數(shù)量積為0,構(gòu)造方程求出t值,得到G點位置;
(3)由
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是平面PAD的法向量,根據(jù)PB與平面ABCD所成的角為45°,求出平面PFD的法向量,代入向量夾角公式,可得答案.
解法二(幾何法)
(I)連接AF,由勾股定理可得DF⊥AF,由PA⊥平面ABCD,由線面垂直性質(zhì)定理可得DF⊥PA,再由線面垂直的判定定理得到DF⊥平面PAF,再由線面垂直的性質(zhì)定理得到PF⊥FD;
(2)過點E作EH∥FD交AD于點H,則EH∥平面PFD,且有AH=
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AD,再過點H作HG∥DP交PA于點G,則HG∥平面PFD且AG=
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AP,由面面平行的判定定理可得平面GEH∥平面PFD,進而由面面平行的性質(zhì)得到EG∥平面PFD.從而確定G點位置;
(Ⅲ)由PA⊥平面ABCD,可得∠PBA是PB與平面ABCD所成的角,即∠PBA=45°,取AD的中點M,則FM⊥AD,F(xiàn)M⊥平面PAD,在平面PAD中,過M作MN⊥PD于N,連接FN,則PD⊥平面FMN,則∠MNF即為二面角A-PD-F的平面角,解三角形MNF可得答案..
試題解析:(1)證明:∵PA⊥平面ABCD,∠BAD=90°,AB=1,AD=2,建立如圖所示的空間直角坐標系A(chǔ)-xyz,則
A(0,0,0),B(1,0,0),F(xiàn)(1,1,0),D(0,2,0).
不妨令P(0,0,t),∵
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=(1,1,-t),
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=(1,-1,0),
∴
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=1×1+1×(-1)+(-t)×0=0,
即PF⊥FD.
(2)解:設平面PFD的法向量為n=(x,y,z),
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由
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得
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令z=1,解得:x=y=
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.
∴n=
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.
設G點坐標為(0,0,m),E
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,則
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,
要使EG∥平面PFD,只需
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·n=0,即
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,得m=
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,從而滿足AG=
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AP的點G即為所求.
(3)解:∵AB⊥平面PAD,∴
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是平面PAD的法向量,易得
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=(1,0,0),又∵PA⊥平面ABCD,∴∠PBA是PB與平面ABCD所成的角,得∠PBA=45°,PA=1,平面PFD的法向量為n=
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.
∴
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.
故所求二面角A-PD-F的余弦值為
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.