摘要:如图5-11.已知平行六面体ABCD-A1B1C1D1的底面ABCD是菱形且∠C1CB=∠C1CD=∠BCD=60°. (1)证明:C1C⊥BD, (2)假定CD=2.CC1=.记面C1BD为α.面CBD为β.求二面角α-BD-β的平面角的余弦值, (3)当的值为多少时.能使A1C⊥平面C1BD?请给出证明.
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