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[几何] 模拟考试一向量题,收集

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realnumber Posted 2014-2-25 13:07 |Read mode
已知$\vv{a},\vv{b}$是平面内两个单位向量,且夹角为60°,若向量$\vv{c}$满足$\vv{c}+\vv{a},\vv{c}+\vv{b}$共线,且方向相反,则$\abs{\vv{c}}$的最小值为________.
答案$\frac{\sqrt{3}}{2}$,几何代数都可以的.

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hongxian Posted 2014-2-25 20:47
$\vv{OA}=\vv{a}$,$\vv{OB}=\vv{b}$,$\vv{c}$的终点放在$O$点,则起点在直线$AB$上

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kuing Posted 2014-2-25 23:59
教师-星变(1664****)  23:48:35
QQ图片20140225235841.jpg

Admin-kuing  23:56:45
画个图就显然了
QQ截图20140225235858.gif
c 的端点在圆上动,于是最小值就是右下角的点到圆的距离,目测大概是B
PS、这题录入的也太懒了,公式中的向量,箭头没有,也没有加粗……

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kuing Posted 2014-2-26 17:34
回复 3# kuing

顺便补充一下这题的代数解法。

设 $\vv b-\vv c=\vv m$,代入条件中,得
\[0=\bigl(\vv a-\vv b+\vv m\bigr)\cdot\bigl(-\vv b+2\vv m\bigr)=2+\bigl(2\vv a-3\vv b\bigr)\cdot\vv m+2\vv m^2,\]
因为
\[\Bigl(\bigl(2\vv a-3\vv b\bigr)\cdot\vv m\Bigr)^2\leqslant\bigl(2\vv a-3\vv b\bigr)^2\vv m^2,\]
由条件容易计算出 $\bigl(2\vv a-3\vv b\bigr)^2=28$,故
\[\bigl(2+2\vv m^2\bigr)^2\leqslant 28\vv m^2,\]
解得
\[\frac{\sqrt7-\sqrt3}2\leqslant \bigl|\vv m\bigr|\leqslant \frac{\sqrt7+\sqrt3}2,\]
当 $2\vv a-3\vv b$ 与 $\vv m$ 共线时取等。

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 Author| realnumber Posted 2014-2-26 18:40
浙江温州汪--(52----46)  18:32:31

QQ图片20140226183924.jpg

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其妙 Posted 2014-2-26 20:03
回复 5# realnumber
浙江省温州市2014届高三第一次适应性考试-17

blog.sina.com.cn/s/blog_879c94820101nayi.html

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kuing Posted 2014-2-26 20:06
回复 6# 其妙

博客dang……

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其妙 Posted 2014-2-26 20:09
回复 7# kuing
有多一个dang了

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