(2014•泰安二模)如图,三棱柱ABC-A1B1C1中,CA=CB,AB=AA1,∠BAA1=60°.
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(2014•泰安二模)如图,三棱柱ABC-A1B1C1中,CA=CB,AB=AA1,∠BAA1=60°.
(Ⅰ)若点M、N分别是边A1B1、BC的中点,求证:MN∥平面ACC1A1
(Ⅱ)证明:AB⊥A1C;
(Ⅲ)若AB=CB=2,A1C=
(Ⅰ)若点M、N分别是边A1B1、BC的中点,求证:MN∥平面ACC1A1
(Ⅱ)证明:AB⊥A1C;
(Ⅲ)若AB=CB=2,A1C=
6 |
(Ⅰ)证明:如图,取AC中点D,连结DA1,DN,
∵N为BC有中点,∴在△ABC中,ND
∥
.
1
2AB,
又∵M为A1B1中点,A1B1∥AB,
∴A1M
∥
.
1
2AB,∴ND
∥
.A1M,
∴四边形A1MND为平行四边形,∴MN∥A1D,
又∵MN不包含于平面ACC1A1,AD1⊂平面ACC1A1,
∴MN∥平面ACC1A1.
(Ⅱ)证明:取AB中点E,连结EC,EA,
在△ABC中,CA=CB,∴EC⊥AB,
又在△AA1B中,AB=AA1,∠BAA1=60°,
∴△AA1B为正三角形,∴A1E⊥AB,
又A1E∩EC=E,
∴AB⊥平面A1EC,
∵A1C⊂平面A1EC,∴AB⊥A1C.
(Ⅲ)∵CA=CB,AB=CB=2,
∴△ABC为边长为2的正三角形,且CE=
3,∴A1E=
3,
又A1C=
6,∴A1E2+CE2=6=A1C2,
∴EC⊥EA1,又EC⊥AB,EA1∩AB=E,
∴以E为原点,EA为x轴,以EA1为y轴,以EC为z轴,建立空间直角坐标系,
∴A(1,0,0),A1(0,
3,0),C(0,0,
∵N为BC有中点,∴在△ABC中,ND
∥
.
1
2AB,
又∵M为A1B1中点,A1B1∥AB,
∴A1M
∥
.
1
2AB,∴ND
∥
.A1M,
∴四边形A1MND为平行四边形,∴MN∥A1D,
又∵MN不包含于平面ACC1A1,AD1⊂平面ACC1A1,
∴MN∥平面ACC1A1.
(Ⅱ)证明:取AB中点E,连结EC,EA,
在△ABC中,CA=CB,∴EC⊥AB,
又在△AA1B中,AB=AA1,∠BAA1=60°,
∴△AA1B为正三角形,∴A1E⊥AB,
又A1E∩EC=E,
∴AB⊥平面A1EC,
∵A1C⊂平面A1EC,∴AB⊥A1C.
(Ⅲ)∵CA=CB,AB=CB=2,
∴△ABC为边长为2的正三角形,且CE=
3,∴A1E=
3,
又A1C=
6,∴A1E2+CE2=6=A1C2,
∴EC⊥EA1,又EC⊥AB,EA1∩AB=E,
∴以E为原点,EA为x轴,以EA1为y轴,以EC为z轴,建立空间直角坐标系,
∴A(1,0,0),A1(0,
3,0),C(0,0,
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