【答案】
分析:(1)以O(shè)為坐標(biāo)原點(diǎn),以AB所在直線為y軸,以O(shè)C所在直線為z軸建立空間直角坐標(biāo)系,求出向量
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與
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的坐標(biāo),利用向量共線的坐標(biāo)表示求證OF∥AC,從而說(shuō)明線面平行;
(2)根據(jù),∠DAB=60°求出D點(diǎn)坐標(biāo),然后求出平面ACD的一個(gè)法向量,找出平面ADB的一個(gè)法向量,利用兩平面法向量所成角的余弦值求解二面角C-AD-B的余弦值;
(3)假設(shè)在
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上存在點(diǎn)G,使得FG∥平面ACD,根據(jù)(1)中的結(jié)論,利用兩面平行的判定定理得到平面OFG∥平面ACD,
從而得到OG∥AD,利用共線向量基本定理得到G的坐標(biāo)(含有參數(shù)),然后由向量
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的模等于圓的半徑求出G點(diǎn)坐標(biāo),最后利用向量
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與平面ACD的法向量所成角的關(guān)系求直線AG與平面ACD所成角的正弦值.
解答:(1)證明:如圖,因?yàn)椤螩AB=45°,連結(jié)OC,則OC⊥AB.
以AB所在的直線為y軸,以O(shè)C所在的直線為z軸,以O(shè)為原點(diǎn),作空間直角坐標(biāo)系O-xyz,
則A(0,-2,0),C(0,0,2).
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,
∵點(diǎn)F為
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的中點(diǎn),∴點(diǎn)F的坐標(biāo)為
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,
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.∴
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,即OF∥AC.
∵OF?平面ACD,AC?平面ACD,∴OF∥平面ACD.
(2)解:∵∠DAB=60°,∴點(diǎn)D的坐標(biāo)
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,
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.
設(shè)二面角C-AD-B的大小為θ,
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為平面ACD的一個(gè)法向量.
由
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有
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即
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取x=1,解得
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,
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.∴
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=
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.
取平面ADB的一個(gè)法向量
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=(0,0,1),
∴
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.
(3)設(shè)在
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上存在點(diǎn)G,使得FG∥平面ACD,∵OF∥平面ACD,∴平面OFG∥平面ACD,則有OG∥AD.
設(shè)
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,∵
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,∴
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.
又∵
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,∴
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,解得λ=±1(舍去-1).∴
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,則G為
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的中點(diǎn).
因此,在
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上存在點(diǎn)G,使得FG∥平面ACD,且點(diǎn)G為
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的中點(diǎn).
設(shè)直線AG與平面ACD所成角為α,∵
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,
根據(jù)(2)的計(jì)算
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為平面ACD的一個(gè)法向量,
∴
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.
因此,直線AG與平面ACD所成角的正弦值為
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.
點(diǎn)評(píng):本題主要考查空間點(diǎn)、線、面位置關(guān)系,線面角、二面角及三角函數(shù)等基礎(chǔ)知識(shí),考查空間想象能力、運(yùn)算能力和推理論證能力,考查用向量方法解決數(shù)學(xué)問(wèn)題的能力,此題是中檔題.