题目
题型：不详难度：来源：
| a |
| b |
| a |
| b |
| m |
| a |
| m |
| b |
| m |
| m |
| A．2 | B．1 | C．
| D．
|
答案
| m |
| a |
| m |
| b |
| m |
| m |
| 0 |
| a |
| b |
| a |
| b |
∵|
| a |
| b |
| a |
| b |
由 (
| a |
| m |
| b |
| m |
| a |
| b |
| m |
| a |
| b |
| m |
| m |
| m |
| a |
| b |
∴|
| m |
| m |
| a |
| b |
| m |
| m |
又∵|
| m |
| m |
故选 B．
核心考点
试题【给定向量a，b且满足|a-b|=1，若对任意向量m满足(a-m)•(b-m)=0，则|m|的最大值与最小值之差为（ ）A．2B．1C．22D．12】；主要考察你对平面向量模和夹角的坐标表示等知识点的理解。[详细]
举一反三
| a |
| b |
| c |
| p |
| ||
|
|
| ||
|
|
| ||
|
|
| p |
| A．[0，1] | B．[0，2] | C．[0，3] | D．[1，2] |
| OA |
| OB |
| AB |
| a |
| b |
| a |
| b |
| b |
①若
| a |
| b |
| a |
| b |
②若
| a |
| b |
| a |
| a |
| b |
| b |
③2|
| b |
| a |
| b |
| b |
| a |
| b |
其中正确的命题序号是______．
| a |
| a |
| b |
| a |
| b |
| 2 |
| b |
A．
| B．3 | C．
| D．2 |