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[几何] 向量最值问题

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hjfmhh Posted 2021-3-11 11:11 |Read mode
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kuing Posted 2021-3-11 15:47
由条件得 `\abs{\bm d}=2` 且 `\abs{2\bm a+\bm c}=\abs{\bm a+2\bm c}`,所以
\begin{align*}
\text{原式}&=2\abs{1-\lambda}+\abs{\bm a+2\bm c}+\abs{2\bm a-2\bm c-2\lambda\bm d}\\
&\geqslant2\abs{1-\lambda}+\abs{3\bm a-2\lambda\bm d}\\
&=2\abs{1-\lambda}+\abs{(3-2\lambda)\bm a+2\sqrt3\lambda\bm b}\\
&=2\abs{1-\lambda}+\sqrt{16\lambda^2-12\lambda+9}\\
&=2\abs{1-\lambda}+\sqrt{\left( \frac14+\frac34 \right)\left( \left( 4\lambda-\frac32 \right)^2+\frac{27}4 \right)}\\
&\geqslant2(1-\lambda)+2\lambda-\frac34+\frac94\\
&=\frac72,
\end{align*}取等懒得算了,略……

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其妙 Posted 2021-3-11 21:20
由条件得 `\abs{\bm d}=2` 且 `\abs{2\bm a+\bm c}=\abs{\bm a+2\bm c}`,所以
\begin{align*}
\text{原式} ...
kuing 发表于 2021-3-11 15:47
我也来玩一下,前几天我在群里发过这道题目的,未见网友解答,我便写了两个解答,其中一个解答和kuing大神居然完全一样!
1.png

不知楼主是否在该群内看见此题的?但是试题样子好像重新编辑过。

现在发出第二个解答!既然是向量题,那么就用纯向量放缩来玩一下!

易知$|\vec d| = |\vec a - \sqrt 3 {\kern 1pt} {\kern 1pt} \vec b| = \sqrt {{{\vec a}^2} + 3{{\vec b}^2}}  = 2$,令$\overrightarrow m  = \vec a + \sqrt 3 {\kern 1pt} {\kern 1pt} \vec b$,则$\vec a \cdot \overrightarrow m  = {\vec a^2}{\text{ + }}\sqrt 3 {\kern 1pt} {\kern 1pt} \vec a \cdot \vec b{\text{ = }}1$,且$\vec d \cdot \overrightarrow m  = {\vec a^2} - 3{\vec b^2} =  - 2$,

则$\left| {2\vec a + \vec c} \right|{\text{ + }}\left| {(1 - \lambda )\vec d} \right|{\text{ + 2}}\left| {\vec a - \vec c - \lambda \vec d} \right|{\text{ = }}\left| {\vec a + 2\vec c} \right|{\text{ + 2}}\left| {\vec a - \vec c - \lambda \vec d} \right|{\text{ + }}\left| {(1 - \lambda )\vec d} \right|$

$ \geqslant \left| {3\vec a - 2\lambda \vec d} \right|{\text{ + }}\left| {(1 - \lambda )\vec d} \right| \geqslant \left| {(3\vec a - 2\lambda \vec d) \cdot \dfrac{1}{2}\overrightarrow m } \right|{\text{ + }}\left| {(1 - \lambda )\vec d} \right|{\text{ = }}\left| {(\dfrac{3}{2}\vec a \cdot \overrightarrow m  - \lambda \vec d \cdot \overrightarrow m )} \right|{\text{ + }}\left| {(1 - \lambda )\vec d} \right|$

${\text{ = }}\left| {\dfrac{3}{2}{\text{ + }}2\lambda } \right|{\text{ + }}\left| {2(1 - \lambda )} \right| \geqslant \dfrac{3}{2} + 2 = \dfrac{7}{2}.$
妙不可言,不明其妙,不着一字,各释其妙!

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isee Posted 2021-3-16 19:40
浙江的题还是上海的?

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facebooker Posted 2021-3-16 21:38
水球卷 浙江的

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