【答案】
分析:(Ⅰ)要證:BD⊥FG,先證BD⊥平面PAC即可.
(Ⅱ)確定點(diǎn)G在線段AC上的位置,使FG∥平面PBD,F(xiàn)G∥平面PBD內(nèi)的一條直線即可.
(Ⅲ)當(dāng)二面角B-PC-D的大小為
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時,求PC與底面ABCD所成角的正切值.
只要作出二面角的平面角,解三角形即可求出結(jié)果.
這三個問題可以利用空間直角坐標(biāo)系,解答(Ⅰ)求數(shù)量積即可.
(Ⅱ)設(shè)才點(diǎn)的坐標(biāo),向量共線即可解答.
(Ⅲ)利用向量數(shù)量積求解法向量,然后轉(zhuǎn)化求出PC與底面ABCD所成角的正切值.
解答:證明:(Ⅰ)∵PA⊥面ABCD,四邊形ABCD是正方形,其對角線BD,AC交于點(diǎn)E,
∴PA⊥BD,AC⊥BD,
∴BD⊥平面PAC,
∵FG?平面PAC,
∴BD⊥FG(5分)
解(Ⅱ):當(dāng)G為EC中點(diǎn),即AG=
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AC時,F(xiàn)G∥平面PBD,(7分)
理由如下:
連接PE,由F為PC中點(diǎn),G為EC中點(diǎn),知FG∥PE,
而FGË平面PBD,PE?平面PBD,
故FG∥平面PBD.(9分)
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解(Ⅲ):作BH^PC于H,連接DH,
∵PA⊥面ABCD,四邊形ABCD是正方形,
∴PB=PD,
又∵BC=DC,PC=PC,
∴△PCB≌△PCD,
∴DH⊥PC,且DH=BH,
∴ÐBHD就是二面角B-PC-D的平面角,(11分)
即ÐBHD=
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,
∵PA⊥面ABCD,∴ÐPCA就是PC與底面ABCD所成的角(12分)
連接EH,則EH⊥BD,ÐBHE=
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,EH⊥PC,
∴tanÐBHE=
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,而BE=EC,
∴
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,∴sinÐPCA=
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,∴tanÐPCA=
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,
∴PC與底面ABCD所成角的正切值是
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(14分)
或用向量方法:
解:以A為原點(diǎn),AB,AD,AP所在的直線分別為x,y,z軸建立空間直角坐標(biāo)系如圖所示,設(shè)正方形ABCD的邊長為1,則A(0,0,0),B(1,0,0),C(1,1,0),D(0,1,0),P(0,0,a)(a>0),E(
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),F(xiàn)(
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),G(m,m,0)(0<m<
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)(2分)
(Ⅰ)
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=(-1,1,0),
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=(
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),
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×
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=-m+
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+m-
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+0=0,
∴BD⊥FG(5分)
(Ⅱ)要使FG∥平面PBD,只需FG∥EP,而
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=(
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),由
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=l
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可得
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,
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解得l=1,m=
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,(7分)
∴G(
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,
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,0),∴
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,
故當(dāng)AG=
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AC時,F(xiàn)G∥平面PBD(9分)
(Ⅲ)設(shè)平面PBC的一個法向量為
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=(x,y,z),
則
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,而
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,
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,
∴
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,取z=1,得
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=(a,0,1),同理可得平面PDC的一個法向量為
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=(0,a,1),
設(shè)
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,
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所成的角為q,則|cosq|=|cos
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|=
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,即
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=
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,∴
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,∴a=1(12分)
∵PA⊥面ABCD,∴ÐPCA就是PC與底面ABCD所成的角,
∴tanÐPCA=
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(14分)
點(diǎn)評:本題考查直線與平面、平面與平面的性質(zhì),空間直線的位置關(guān)系,空間直角坐標(biāo)系,空間想象能力,邏輯思維能力,是難度較大題目.