题目内容
如图,直棱柱(侧棱垂直于底面的棱柱)ABC-A1B1C1,在底面ABC中,CA=CB=1,∠BCA=90°,棱AA1=2,M,N分别为A1B1,A1A的中点.
(1)求cos<
,
>的值;
(2)求证:BN⊥平面C1MN.

(1)求cos<
| BA1 |
| CB1 |
(2)求证:BN⊥平面C1MN.
以C为原点,CA,CB,CC1所在直线分别为x轴,y轴,z轴,建立如图所示的坐标系C-xyz,
(1)依题意,A1(1,0,2),C(0,0,0),B1(0,1,2),B(0,1,0),
∴
=(1,-1,2),
=(0,1,2),
∴
•
=1×0+(-1)×1+2×2=3,
又|
|=
,|
|=
,
∴cos<
,
>=
=
…6分
证明:(2)A1(1,0,2),C1(0,0,2),B1(0,1,2),N(1,0,1),
∴M(
,
,2),∴
=(
,
,2),
=(1,0,-1),
=(1,-1,1),
∴
•
=
×1+
×(-1)+1×0=0,同理可求
•
=0,
∴
⊥
,
⊥
,C1M∩C1N=C1,
∴BN⊥平面C1MN…12分.
(1)依题意,A1(1,0,2),C(0,0,0),B1(0,1,2),B(0,1,0),
∴
| BA1 |
| CB1 |
∴
| BA1 |
| CB1 |
又|
| BA1 |
| 6 |
| CB1 |
| 5 |
∴cos<
| BA1 |
| CB1 |
| ||||
|
|
| ||
| 10 |
证明:(2)A1(1,0,2),C1(0,0,2),B1(0,1,2),N(1,0,1),
∴M(
| 1 |
| 2 |
| 1 |
| 2 |
| C1M |
| 1 |
| 2 |
| 1 |
| 2 |
| C1N |
| BN |
∴
| C1M |
| BN |
| 1 |
| 2 |
| 1 |
| 2 |
| C1N |
| BN |
∴
| C1M |
| BN |
| C1N |
| BN |
∴BN⊥平面C1MN…12分.
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