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[几何] 立体几何中动点问题

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lrh2006 Posted at 2022-11-26 10:49:35 |Read mode
Last edited by hbghlyj at 2025-3-9 01:21:23(多选题) 如图,等腰直角△ABC中,AC=CB,点P为平面ABC外一动点,
满足PB=AB=2,$ \angle PBA=\frac{\pi}{2}$,则存在点P使得( )
A. $AB\perp PC$     
B. PB与平面PAC所成角为$\frac{\pi}{4}$
C. $PC=\frac{16}{5}$
D. 二面角P-AC-B的大小为$\frac{\pi}{3}$

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2025-4-20 22:20 GMT+8

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