题目内容
已知|| OA |
| OB |
| 3 |
| OA |
| OB |
| OC |
| OA |
| OB |
| m |
| n |
分析:先根据
•
=0,可得
⊥
,又因为
•
=OC×
cos60°=
=
×
|OC |
•
=|OC|×1×cos30°=
|OC|=1×
|OC|,所以可得:
在x轴方向上的分量为
|OC|
在y轴方向上的分量为
|OC|,又根据
=m
+n
=
n
+m
,可得答案.
| OA |
| OB |
| OA |
| OB |
| OC |
| OB |
| 3 |
| ||
| 2 |
| OC |
| 3 |
| 1 |
| 2 |
| OC |
| OA |
| ||
| 2 |
| ||
| 2 |
| OC |
| 1 |
| 2 |
| OC |
| ||
| 2 |
| OC |
| OA |
| OB |
| 3 |
| i |
| j |
解答:解:∵|
|=1,|
|=
,
•
=0,
⊥
•
=OC×
cos60°=
=
×
|OC |
•
=|OC|×1×cos30°=
|OC|=1×
|OC|
∴
在x轴方向上的分量为
|OC|
在y轴方向上的分量为
|OC|
∵
=m
+n
=
n
+m
∴
|OC|=
n,
|OC|=m
两式相比可得:
=3.
故答案为:3
| OA |
| OB |
| 3 |
| OA |
| OB |
| OA |
| OB |
| OC |
| OB |
| 3 |
| ||
| 2 |
| OC |
| 3 |
| 1 |
| 2 |
| OC |
| OA |
| ||
| 2 |
| ||
| 2 |
∴
| OC |
| 1 |
| 2 |
| OC |
| ||
| 2 |
∵
| OC |
| OA |
| OB |
| 3 |
| i |
| j |
∴
| 1 |
| 2 |
| 3 |
| ||
| 2 |
两式相比可得:
| m |
| n |
故答案为:3
点评:本题主要考查向量数量积的几何意义.对于向量数量积要明确其几何意义和运算法则.
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