题目内容
已知|
|=1,|
|=1,∠AOB=
,
=
+
,则
与
的夹角大小为 .
| OA |
| OB |
| 2π |
| 3 |
| OC |
| 1 |
| 2 |
| OA |
| 1 |
| 4 |
| OB |
| OA |
| OC |
考点:平面向量数量积的运算
专题:计算题
分析:利用向量夹角公式计算cosθ=
,再利用特殊角的三角函数值确定夹角.
| ||||
|
|
解答:
解:∵
=
+
,∴|
|2=|
+
|2=
2+
•
+
2=
×12+
×1×1×cos
+
×12=
∴|
|=
,
又
•
=(
+
)•
=
2+
•
=
,
∴
与
的夹角的夹角θ的余弦值为cosθ=
=
=
∴θ=
故答案为:
| OC |
| 1 |
| 2 |
| OA |
| 1 |
| 4 |
| OB |
| OC |
| 1 |
| 2 |
| OA |
| 1 |
| 4 |
| OB |
| 1 |
| 4 |
| OA |
| 1 |
| 4 |
| OB |
| OA |
| 1 |
| 16 |
| OB |
| 1 |
| 4 |
| 1 |
| 4 |
| 2π |
| 3 |
| 1 |
| 16 |
| 3 |
| 16 |
∴|
| OC |
| ||
| 4 |
又
| OA |
| OC |
| 1 |
| 2 |
| OA |
| 1 |
| 4 |
| OB |
| OA |
| 1 |
| 2 |
| OA |
| 1 |
| 4 |
| OB |
| OA |
| 3 |
| 8 |
∴
| OA |
| OC |
| ||||
|
|
| ||||
1×
|
| ||
| 2 |
∴θ=
| π |
| 6 |
故答案为:
| π |
| 6 |
点评:本题考查向量夹角的计算,牢记公式,准确计算为要.
练习册系列答案
相关题目
下列关于向量的等式中,正确的是( )
A、
| ||||||||
B、
| ||||||||
C、
| ||||||||
D、
|