题目内容
如图,已知正三棱柱ABC-A1B1C1的各条棱长都相等,则异面直线AB1和A1C所成的角的余弦值大小是______.
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延长CA到D,使得AD=AC,则ADA1B1为平行四边形,
∴AB1∥A1D,
∴∠DA1C就是异面直线AB1和A1C所成的角,
又三角形ABC为等边三角形,设AB=AA1=1,∠CAD=120°
则CD=
=
;A1C=A1D=
,
在△A1CD中,cos∠DA1C=
=
.
故答案是:
.
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∴AB1∥A1D,
∴∠DA1C就是异面直线AB1和A1C所成的角,
又三角形ABC为等边三角形,设AB=AA1=1,∠CAD=120°
则CD=
1+1-2×1×1×(-
|
3 |
2 |
在△A1CD中,cos∠DA1C=
2+2-3 | ||||
2×
|
1 |
4 |
故答案是:
1 |
4 |
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