解法一:(1)∵AC
1是正方體,∴AD⊥面DC
1.
又D
1F?面DC
1,∴AD⊥D
1F.
(2)取AB中點G,連接A
1G,F(xiàn)G.
因為F是CD的中點,所以GF、AD平行且相等,
又A
1D
1、AD平行且相等,所以GF、A
1D
1平行且相等,故GFD
1A
1是平行四邊形,A
1G∥D
1F.
設(shè)A
1G與AE相交于點H,則∠AHA
1是AE與D
1F所成的角,
因為E是BB
1的中點,所以Rt△A
1AG≌Rt△ABE,
∴∠GA
1A=∠GAH,從而∠AHA
1=90°,即直線AE與D
1F所成角為直角.
(3)由(1)知AD⊥D
1F,由(2)知AE⊥D
1F,
又AD∩AE=A,所以D
1F⊥面AED.
又因為D
1F?面A
1FD
1,
所以面AED⊥面A
1FD
1.
(4)連接GE,GD
1.
∵FG∥A
1D
1,∴FG∥面A
1ED
1,
∵AA
1=2,
∴面積S
△A1GE=S
ABB1A1-2S
△A1AG-S
△GBE=
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又
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=
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∴
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解法二:利用用向量求解
解:設(shè)正方體的棱長為2,以D為原點,DA為x軸,DC為y軸,DD
1為z軸建立空間直角坐標(biāo)系,
則D(0,0,0),A(2,0,0),F(xiàn)(0,1,0),E(2,2,1),A
1(2,0,2),D
1(0,0,2),
(1)∵
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,
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,得
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,∴AD⊥D
1F;
(2)又
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,得
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=
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∴AE與D
1F所成的角為90°
(3)由題意:
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,
設(shè)平面AED的法向量為
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,設(shè)平面A
1FD
1的法向量為
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,
由
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由
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得
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=
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∴面AED⊥面A
1FD
1.
(4)∵AA
1=2,
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,
平面A
1FD
1的法向量為
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=
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,
∴E到平面A
1FD
1的距離
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=
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,
∴
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.
分析:解法一:傳統(tǒng)證法.(1)利用線面垂直,證明線線垂直;
(2)設(shè)A
1G與AE相交于點H,先證∠AHA
1是AE與D
1F所成的角,再求直線AE與D
1F所成角;
(3)利用線面垂直,證明面面垂直;
(4)利用轉(zhuǎn)換底面的方法,求三棱錐的體積;
解法二:向量證法.設(shè)正方體的棱長為2,以D為原點,DA為x軸,DC為y軸,DD
1為z軸建立空間直角坐標(biāo)系,
則D(0,0,0),A(2,0,0),F(xiàn)(0,1,0),E(2,2,1),A
1(2,0,2),D
1(0,0,2),
(1)利用
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,可證AD⊥D
1F;
(2)求得
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=
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,可求AE與D
1F所成的角;(3)由題意:

,
設(shè)平面AED的法向量為
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,設(shè)平面A
1FD
1的法向量為
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,證明平面的法向量垂直,即可證明面AED⊥面A
1FD
1.
(4)先求得
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=
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,計算E到平面A
1FD
1的距離
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=
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,即可求三棱錐的體積.
點評:本題重點考查線面垂直、面面垂直,考查三棱錐的體積,兩法并用,注意比較,細(xì)細(xì)體會.