摘要:又C1C平面AC1.∴C1C⊥BD.知AC⊥BD.C1O⊥BD.∴∠C1OC是二面角α―BD―β的平面角.
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如图,已知平行六面体ABCD-A1B1C1D1的底面ABCD上菱形,且∠C1CB=∠C1CD=∠BCD,
(1)证明:C1C⊥BD;
(2)当
的值为多少时,能使A1C⊥平面C1BD?请给出证明.
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(1)证明:C1C⊥BD;
(2)当
CD | CC1 |
如图,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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(1)证明:C1C⊥BD;
(2)假定CD=2,CC1=
3 |
2 |
(3)当
CD |
CC1 |
如图,已知斜四棱柱ABCD-A1B1C1D1的底面ABCD是菱形,且∠C1CB=∠C1CD=∠BCD.
(1) 证明:C1C⊥BD;
(2) 当的值为多少时,能使A1C⊥平面C1BD?请给出证明
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