【答案】
分析:法一:(1)要證AB
1⊥面A
1BD,只需證明直線AB
1垂直面A
1BD內(nèi)的兩條相交直線B
1O、AB
1即可;
(2)設(shè)AB
1與A
1B交于點G,在平面A
1BD中,作GF⊥A
1D于F,連接AF,
說明∠AFG為二面角A-A
1D-B的平面角,然后解三角形,求二面角A-A
1D-B的大小;
(3)利用等體積法

,求點C到平面A
1BD的距離.
法二:建立空間直角坐標系,求出相關(guān)向量,利用向量的數(shù)量積等于0證明垂直,
(1)求證:AB
1⊥面A
1BD;
向量共線證明平行,向量數(shù)量積求出二面角的大小
(2)求二面角A-A
1D-B的大��;
距離公式求出距離,解答(3)求點C到平面A
1BD的距離.
解答:
證明:法一:(Ⅰ)取BC中點O,連接AO.∵△ABC為正三角形,∴AO⊥BC.
∵正三棱柱ABC-A
1B
1C
1中,平面ABC⊥平面BCC
1B
1,∴AO⊥平面BCC
1B
1.
連接B
1O,在正方形BB
1C
1C中,O,D分別為BC,CC
1的中點,∴B
1O⊥BD,∴AB
1⊥BD.
在正方形ABB
1A
1中,AB
1⊥A
1B,∴AB
1⊥平面A
1BD.
(Ⅱ)設(shè)AB
1與A
1B交于點G,在平面A
1BD中,作GF⊥A
1D于F,連接AF,
由(Ⅰ)得AB
1⊥平面A
1BD.∴AF⊥A
1D,∴∠AFG為二面角A-A
1D-B的平面角.
在△AA
1D中,由等面積法可求得

,
又∵

,
∴

.
所以二面角A-A
1D-B的大小為

.
(Ⅲ)△A
1BD中,

,S
△BCD=1.
在正三棱柱中,A
1到平面BCC
1B
1的距離為=

.
設(shè)點C到平面A
1BD的距離為d.
由

得
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,∴

.∴點C到平面C的距離為

.
法二:(Ⅰ)取BC中點O,連接AO.
∵△ABC為正三角形,
∴AO⊥BC.
∵在正三棱柱ABC-A
1B
1C
1中,平面ABC⊥平面BCC
1B
1,
∴AO⊥平面BCC
1B
1.
取B
1C
1中點O
1,以O(shè)為原點,

,

,

的方向為x,y,z軸的正方向建立空間直角坐標系,
則B(1,0,0),D(-1,1,0),

,

,B
1(1,2,0),
∴
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,

,

.
∵

,

,
∴

,

.
∴AB
1⊥平面A
1BD.

(Ⅱ)設(shè)平面A
1AD的法向量為n=(x,y,z).
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,

.
∵
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,
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,
∴
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∴

∴
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令z=1得
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為平面A
1AD的一個法向量.
由(Ⅰ)知AB
1⊥平面A
1BD,∴

為平面A
1BD的法向量.cos<n,

.
∴二面角A-A
1D-B的大小為

.
(Ⅲ)由(Ⅱ),

為平面A
1BD法向量,∵

.
∴點C到平面A
1BD的距離

.
點評:本小題主要考查直線與平面的位置關(guān)系,二面角的大小,點到平面的距離等知識,考查空間想象能力、邏輯思維能力和運算能力.