题目内容
| OA |
| OB |
| 2π |
| 3 |
(1)|
| OA |
| OB |
(2)
| AB |
| AC |
分析:(1)根据条件先求出
•
的值,再求出|
+
|=
的值;
(2)根据条件求出
•
,
•
的值,再由减法运算得
•
=(
-
)•(
-
),再展开进行求解即可.
| OA |
| OB |
| OA |
| OB |
(
|
(2)根据条件求出
| OA |
| OC |
| OB |
| OC |
| AB |
| AC |
| OB |
| OA |
| OC |
| OA |
解答:解:(1)∵
和
的长度为1,夹角为
,
∴
•
=|
||
|cos
=-
,
∴|
+
|=
=
=1,
(2)∵点C是以O为圆心的劣弧AB的中点,
∴∠AOC=∠BOC=
,∴
•
=
•
=
,
∴
•
=(
-
)•(
-
)
=
•
-
•
-
•
+
•
=
-(-
)-
+1=
.
| OA |
| OB |
| 2π |
| 3 |
∴
| OA |
| OB |
| OA |
| OB |
| 2π |
| 3 |
| 1 |
| 2 |
∴|
| OA |
| OB |
(
|
|
(2)∵点C是以O为圆心的劣弧AB的中点,
∴∠AOC=∠BOC=
| π |
| 3 |
| OA |
| OC |
| OB |
| OC |
| 1 |
| 2 |
∴
| AB |
| AC |
| OB |
| OA |
| OC |
| OA |
=
| OB |
| OC |
| OB |
| OA |
| OA |
| OC |
| OA |
| OA |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
点评:本题考查了向量的数量积和减法运算,主要利用定义和性质进行求解.
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