题目内容
正方体ABCD-A1B1C1D1中二面角A1-BD-C1的余弦值为______.
如图所示,取BD的中点O,连接A1O,C1O,则A1O⊥BD,C1O⊥BD,
∴∠A1OC1为二面角A1-BD-C1的平面角
设正方体的棱长为1,则A1C1=
,A1O=C1O=
,
∴cos∠A1OC1=
=-
故答案为:-
∴∠A1OC1为二面角A1-BD-C1的平面角
设正方体的棱长为1,则A1C1=
2 |
| ||
2 |
∴cos∠A1OC1=
| ||||||||
2•
|
1 |
3 |
故答案为:-
1 |
3 |
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