题目内容
正方体ABCD-A1B1C1D1中,E,F分别是棱AA1,BB1的中点.
(1)求证:平面A1BC1∥平面ACD1;
(2)求异面直线A1F与D1E所成的角的余弦值.
(1)求证:平面A1BC1∥平面ACD1;
(2)求异面直线A1F与D1E所成的角的余弦值.
证明:(1)如图,
连结AC,AD1,CD1,A1C1,A1B,C1B.
∵ABCD-A1B1C1D1是正方体,∴AA1∥CC1,AA1=CC1,
∴四边形AA1C1C为平行四边形,∴A1C1∥AC.
A1C1?平面ACD1,AC?平面ACD1,∴A1C1∥平面ACD1;
∵A1D1∥BC,A1D1=BC,∴四边形A1BCD1为平行四边形,∴A1B∥CD1.
A1B?平面ACD1,CD1?平面ACD1,∴A1B∥?平面ACD1,
又A1B∩A1C1=A1,
∴平面A1BC1∥平面ACD1;
(2)连结C1F,∵E,F分别是棱AA1,BB1的中点,∴EF∥C1D1,EF=C1D1
∴EFC1D1是平行四边形,∴D1F∥C1E.
设正方体ABCD-A1B1C1D1的棱长为2,解直角三角形求得A1C1=2
,A1F=C1F=
.
在△A1C1F中,由余弦定理得cos∠A1FC1=
=
=
.
∴异面直线A1F与D1E所成的角的余弦值是
.
连结AC,AD1,CD1,A1C1,A1B,C1B.
∵ABCD-A1B1C1D1是正方体,∴AA1∥CC1,AA1=CC1,
∴四边形AA1C1C为平行四边形,∴A1C1∥AC.
A1C1?平面ACD1,AC?平面ACD1,∴A1C1∥平面ACD1;
∵A1D1∥BC,A1D1=BC,∴四边形A1BCD1为平行四边形,∴A1B∥CD1.
A1B?平面ACD1,CD1?平面ACD1,∴A1B∥?平面ACD1,
又A1B∩A1C1=A1,
∴平面A1BC1∥平面ACD1;
(2)连结C1F,∵E,F分别是棱AA1,BB1的中点,∴EF∥C1D1,EF=C1D1
∴EFC1D1是平行四边形,∴D1F∥C1E.
设正方体ABCD-A1B1C1D1的棱长为2,解直角三角形求得A1C1=2
2 |
5 |
在△A1C1F中,由余弦定理得cos∠A1FC1=
A1F2+C1F2-A1C12 |
2A1F•C1F |
(
| ||||||
2×
|
1 |
5 |
∴异面直线A1F与D1E所成的角的余弦值是
1 |
5 |
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