试题分析:(1)连接AC,交BQ于N,连接MN,在三角形PAC中,利用中位线定理证明PA//MN,由线线平行得线面平行;(2)证PQ⊥AD,QB⊥AD,由PQ∩BQ=Q,所以AD⊥平面PBQ,再利用线面垂直得面面垂直;(3)先证PQ⊥面ABCD,(注意此步不可省略),再以Q为原点建立空间直角坐标系,写出各点坐标及平面BQC的法向量
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,并设
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,利用关系PM=tMC,用坐标表示出来,列方程解出
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,并得
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,
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,从而易得平面MBQ法向量为
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,再由数量积运算得
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,可得t值.
试题解析:证明:(1)连接AC,交BQ于N,连接MN. 1分
∵BC∥AD且BC=
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AD,即BC
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AQ.∴四边形BCQA为平行四边形,且N为AC中点,
又∵点M是棱PC的中点,∴ MN // PA 2分
∵ MN
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平面MQB,PA
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平面MQB, 3分
∴ PA // 平面MBQ. 4分
(2)∵AD // BC,BC=
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AD,Q为AD的中点,∴四边形BCDQ为平行四边形,∴CD // BQ . 6分
∵∠ADC=90° ∴∠AQB=90° 即QB⊥AD.
又∵平面PAD⊥平面ABCD,且平面PAD∩平面ABCD=AD, 7分
∴BQ⊥平面PAD. 8分
∵BQ
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平面PQB,∴平面PQB⊥平面PAD. 9分
另证:AD // BC,BC=
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AD,Q为AD的中点∴ BC // DQ 且BC= DQ,
∴ 四边形BCDQ为平行四边形,∴CD // BQ .
∵ ∠ADC=90° ∴∠AQB=90° 即QB⊥AD. 6分
∵ PA=PD, ∴PQ⊥AD. 7分
∵ PQ∩BQ=Q,∴AD⊥平面PBQ. 8分
∵ AD
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平面PAD,∴平面PQB⊥平面PAD. 9分
(Ⅲ)∵PA=PD,Q为AD的中点, ∴PQ⊥AD.
∵平面PAD⊥平面ABCD,且平面PAD∩平面ABCD=AD, ∴PQ⊥平面ABCD. 10分
(不证明PQ⊥平面ABCD直接建系扣1分)
如图,以Q为原点建立空间直角坐标系.
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则平面BQC的法向量为
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;
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,
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,
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,
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. 11分
设
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,
则
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,
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,∵
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,
∴
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, ∴
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, 12分
在平面MBQ中,
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,
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,
∴ 平面MBQ法向量为
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. 13分
∵二面角M-BQ-C为30°,
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,∴
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. 14分