(2014?太原一模)如图,在斜三棱柱ABC-A1B1C1中,点O是A1C1的中点,AO⊥平面A1B1C1.已知∠BCA=90°,AA

2025-05-13 05:53:34
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回答(1):

(1)证明:∵AO⊥平面A1B1C1
∴AO⊥B1C1 
又∵A1C1⊥B1C1,且A1C1∩AO=0,
∴B1C1⊥平面A1C1CA,∴A1C⊥B1C1
又∵AA1=AC,
∴四边形A1C1CA为菱形,
∴A1C⊥AC1,且B1C1∩AC1=C1
∴A1C⊥平面AB1C1
∴AB1⊥A1C.
(2)∵CC1∥平面AA1B1
∴点C到平面AA1B2的距离与点C1到平面AA1B1的距离相等,
设C1到平面AA1B1的距离为d,
VA?A1B1C1=VC1?AA1B1

1
3
?
1
2
?A1C1?B1C1?AO=
1
3
?S△AA1B1?d,
又∵在△AA1B1中,A1B1=AB1=2
2
AA1=2,S△AA1B1
7

∴d=
2
21
7