摘要:∴DF=C1H=2. (Ⅱ)延长C1E与CB交于G.连AG.则平面AEC1F与平面ABCD相交于AG.过C作CM⊥AG.垂足为M.连C1M.由三垂线定理可知AG⊥C1M.由于AG⊥面C1MC.
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(1)求椭圆的离心率;
(2)求∠F1QF2的范围;
(3)当QF2⊥AB时,延长QF2与椭圆交于另一点P,若△F1PQ的面积为20
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(1)求证GA∥平面FMC;
(2)求直线DM与平面ABEF所成角. 查看习题详情和答案>>