题目内容
若|
+
|=|
-
|=2|
|,则向量
+
与
的夹角为( )
| a |
| b |
| a |
| b |
| a |
| a |
| b |
| a |
A.
| B.
| C.
| D.
|
∵|
+
|=|
-
|=2|
|,
∴|
+
|2=|
-
|2,两边平方
可得
2+2
•
+
2=
2-2
•
+
2,
化简可得
•
=0,
设向量
+
与
的夹角为θ
则可得cosθ=
=
=
=
,又θ∈[0,π],故θ=
故选B
| a |
| b |
| a |
| b |
| a |
∴|
| a |
| b |
| a |
| b |
可得
| a |
| a |
| b |
| b |
| a |
| a |
| b |
| b |
化简可得
| a |
| b |
设向量
| a |
| b |
| a |
则可得cosθ=
(
| ||||||
|
|
| ||||||
|
|
=
|
| ||
2|
|
| 1 |
| 2 |
| π |
| 3 |
故选B
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