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(2014•揭阳二模)如图,已知三棱柱ABC-A1B1C1的侧棱与底面垂直,且∠ACB=90°,∠BAC=30°,BC=1,AA1=6,点P、M、N分别为BC1、CC1、AB1的中点.(1)求证:PN∥平面ABC;(2)求证:A1M⊥AB1C1;(3)
题目详情
(2014•揭阳二模)如图,已知三棱柱ABC-A1B1C1的侧棱与底面垂直,且∠ACB=90°,∠BAC=30°,BC=1,AA1=
,点P、M、N分别为BC1、CC1、AB1的中点.
(1)求证:PN∥平面ABC;
(2)求证:A1M⊥AB1C1;
(3)求点M到平面AA1B1的距离.
6 |
(1)求证:PN∥平面ABC;
(2)求证:A1M⊥AB1C1;
(3)求点M到平面AA1B1的距离.
▼优质解答
答案和解析
(1)证明:连结CB1,
∵P是BC1的中点,∴CB1过点P,--(1分)
∵N为AB1的中点,∴PN∥AC,---------------------------(2分)
∵AC⊂面ABC,PN⊄面ABC,
∴PN∥平面ABC.--------------------------------------(4分)
(2)证法一:连结AC1,在直角△ABC中,
∵BC=1,∠BAC=30°,
∴AC=A1C1=
-----------------------------------(5分)
∵
=
=
,
∴Rt△A1C1M~Rt△C1CA--------------------------------------------------------(7分)
∴∠A1MC1=∠CAC1,∴∠AC1C+∠CAC1=∠AC1C+∠A1MC1=90°
∴AC1⊥A1M.-------------------------------------------------------------------(8分)
∵B1C1⊥C1A1,CC1⊥B1C1,且C1A1∩CC1=C1
∴B1C1⊥平面AA1CC1,-----------------------------------------------------------(9分)
∴B1C1⊥A1M,又AC1∩B1C1=C1,故A1M⊥平面A B1C1,-------------------------(11分)
证法二:连结AC1,在直角△ABC中,∵BC=1,∠BAC=30°,
∴AC=A1C1=
-------------------------------------------------------------(5分)
设∠AC1A1=α,∠MA1C1=β
∵tanαtanβ=
•
=
∵P是BC1的中点,∴CB1过点P,--(1分)
∵N为AB1的中点,∴PN∥AC,---------------------------(2分)
∵AC⊂面ABC,PN⊄面ABC,
∴PN∥平面ABC.--------------------------------------(4分)
(2)证法一:连结AC1,在直角△ABC中,
∵BC=1,∠BAC=30°,
∴AC=A1C1=
3 |
∵
CC1 |
A1C1 |
A1C1 |
MC1 |
2 |
∴Rt△A1C1M~Rt△C1CA--------------------------------------------------------(7分)
∴∠A1MC1=∠CAC1,∴∠AC1C+∠CAC1=∠AC1C+∠A1MC1=90°
∴AC1⊥A1M.-------------------------------------------------------------------(8分)
∵B1C1⊥C1A1,CC1⊥B1C1,且C1A1∩CC1=C1
∴B1C1⊥平面AA1CC1,-----------------------------------------------------------(9分)
∴B1C1⊥A1M,又AC1∩B1C1=C1,故A1M⊥平面A B1C1,-------------------------(11分)
证法二:连结AC1,在直角△ABC中,∵BC=1,∠BAC=30°,
∴AC=A1C1=
3 |
设∠AC1A1=α,∠MA1C1=β
∵tanαtanβ=
AA1 |
A1C1 |
MC1 |
A1C1 |
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