题目内容
(2011•静海县一模)已知
=(2,0),
=(2,2),
=(2,1),则
与
夹角的正弦值为
.
| OB |
| OC |
| CA |
| OA |
| OB |
| 3 |
| 5 |
| 3 |
| 5 |
分析:设
=(x,y),则由
=
-
=(2,1),求得x、y的值,可得
的坐标,再利用两个向量的夹角公式求得
与
夹角的余弦值,从而求得则
与
夹角的正弦值.
| OA |
| CA |
| OA |
| OC |
| OA |
| OA |
| OB |
| OA |
与
| OB |
解答:解:设
=(x,y),则由
=
-
=(x-2,y-2)=(2,1),可得
,即
=(4,3),
∴cos<
,
>=
=
=
,故sin<
,
>=
,
故答案为
.
| OA |
| CA |
| OA |
| OC |
|
| OA |
∴cos<
| OA |
| OB |
| ||||
|
|
| 8 |
| 5×2 |
| 4 |
| 5 |
| OA |
| OB |
| 3 |
| 5 |
故答案为
| 3 |
| 5 |
点评:本题主要考查两个向量坐标形式的运算,两个向量的夹角公式的应用,属于中档题.
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