题目内容
6.已知a、b满足|a|=1,|b|=$\sqrt{5}$,|a+b|=|a-b|,则|2a-b|=3.分析 根据题意得|$\overrightarrow{a}+\overrightarrow{b}$|2=|$\overrightarrow{a}-\overrightarrow{b}$|2,展开即可得到$\overrightarrow{a}•\overrightarrow{b}$的值,再计算$\sqrt{{|\begin{array}{l}{2\overrightarrow{a}-\overrightarrow{b}}\end{array}|}^{2}}$即可.
解答 解:∵|$\overrightarrow{a}+\overrightarrow{b}$|=|$\overrightarrow{a}-\overrightarrow{b}$|,
∴|$\overrightarrow{a}+\overrightarrow{b}$|2=|$\overrightarrow{a}-\overrightarrow{b}$|2,
即${\overrightarrow{a}}^{2}+2\overrightarrow{a}•\overrightarrow{b}+{\overrightarrow{b}}^{2}={\overrightarrow{a}}^{2}-2\overrightarrow{a}•\overrightarrow{b}+{\overrightarrow{b}}^{2}$,
化简得$\overrightarrow{a}•\overrightarrow{b}=0$,
所以$|\begin{array}{l}{2\overrightarrow{a}-\overrightarrow{b}}\end{array}|$=$\sqrt{{|\begin{array}{l}{2\overrightarrow{a}-\overrightarrow{b}}\end{array}|}^{2}}$
=$\sqrt{4{\overrightarrow{a}}^{2}-4\overrightarrow{a}•\overrightarrow{b}+{\overrightarrow{b}}^{2}}$
=$\sqrt{4+5}$
=3,
故答案为:3.
点评 本题考查数量积的运算性质,属基础题.
| A. | x2f(lnx1)<x1f(lnx2) | B. | x2f(lnx1)>x2f(lnx2) | C. | x1f(lnx1)>x2f(lnx2) | D. | x1f(lnx1)<x2f(lnx2) |