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2021年高考数学天津15

  2022-05-03 08:25:41  

15.(5分)在边长为1的等边三角形$ABC$中,$D$为线段$BC$上的动点,$DE\bot AB$且交$AB$于点$E$,$DF//AB$且交$AC$于点$F$,则$\vert 2\overrightarrow{BE}+\overrightarrow{DF}\vert$的值为____;$(\overrightarrow{DE}+\overrightarrow{DF})\cdot \overrightarrow{DA}$的最小值为 ____.
分析:设$BE=x$,表示出$BD=2x$,$DE=\sqrt{3}x$,$DC=1-2x$,利用数量积的定义与性质即可求出.
解:如图,设$BE=x$,

$\because \Delta ABC$是边长为1等边三角形,$DE\bot AB$,
$\therefore \angle BDE=30\circ$,$BD=2x$,$DE=\sqrt{3}x$,$DC=1-2x$,
$\because DF//AB$,$\therefore \Delta DFC$是边长为$1-2x$等边三角形,$DE\bot DF$,
$\therefore (2\overrightarrow{BE}+\overrightarrow{DF})^{2}=4{\overrightarrow{BE}}^{2}+4\overrightarrow{BE}\cdot \overrightarrow{DF}+{\overrightarrow{DF}}^{2}=4x^{2}+4x(1-2x)\times \cos 0\circ +(1-2x)^{2}=1$,
则$\vert 2\overrightarrow{BE}+\overrightarrow{DF}\vert =1$,
$\because (\overrightarrow{DE}+\overrightarrow{DF})\cdot \overrightarrow{DA}=(\overrightarrow{DE}+\overrightarrow{DF})\cdot (\overrightarrow{DE}+\overrightarrow{EA})={\overrightarrow{DE}}^{2}+\overrightarrow{DF}\cdot \overrightarrow{EA}$
$={(\sqrt{3}x)}^{2}+(1-2x)\times (1-x)=5x^{2}-3x+1$
$=5{(x-\dfrac{3}{10})}^{2}+\dfrac{11}{20}$,$x\in (0,\dfrac{1}{2})$,
$\therefore (\overrightarrow{DE}+\overrightarrow{DF})\cdot \overrightarrow{DA}$的最小值为$\dfrac{11}{20}$.
故答案为:1,$\dfrac{11}{20}$.
点评:本题考查向量的数量积的定义,向量的运算法则,二次函数求最值,属于中档题.

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