题目内容
三棱柱ABC-A1B1C1中,底面边长和侧棱长都相等,∠BAA1=∠CAA1=60°,则异面直线AB1与BC1所成角的余弦值为
.
| ||
| 6 |
| ||
| 6 |
分析:先选一组基底,再利用向量加法和减法的三角形法则和平行四边形法则将两条异面直线的方向向量用基底表示,最后利用夹角公式求异面直线AB1与BC1所成角的余弦值即可
解答:解:如图,
设
=
,
=
,
=
,棱长均为1,
则
•
=
,
•
=
,
•
=
∵
=
+
,
=
+
=
-
+
∴
•
=(
+
)•(
-
+
)=
•
-
2+
•
+
•
-
•
+
2
=
•
-
2+
•
+
2=
-1+
+1=1
|
|=
=
=
|
|=
=
=
∴cos<
,
>=
=
=
∴异面直线AB1与BC1所成角的余弦值为
| AA1 |
| c |
| AB |
| a |
| AC |
| b |
则
| a |
| b |
| 1 |
| 2 |
| b |
| c |
| 1 |
| 2 |
| a |
| c |
| 1 |
| 2 |
∵
| AB1 |
| a |
| c |
| BC1 |
| BC |
| BB1 |
| b |
| a |
| c |
∴
| AB1 |
| BC1 |
| a |
| c |
| b |
| a |
| c |
| a |
| b |
| a |
| a |
| c |
| b |
| c |
| a |
| c |
| c |
=
| a |
| b |
| a |
| b |
| c |
| c |
| 1 |
| 2 |
| 1 |
| 2 |
|
| AB1 |
(
|
| 1+1+1 |
| 3 |
|
| BC1 |
(
|
| 1+1+1-1-1+1 |
| 2 |
∴cos<
| AB1 |
| BC1 |
| ||||
|
|
| 1 | ||||
|
| ||
| 6 |
∴异面直线AB1与BC1所成角的余弦值为
| ||
| 6 |
点评:本题主要考查了空间向量在解决立体几何问题中的应用,空间向量基本定理,向量数量积运算的性质及夹角公式的应用,有一定的运算量
练习册系列答案
相关题目