已知正方形 A B C D 的边长为 1，当每个 λ_ i…——2019 高考数学第 17 题答案解析

【答案】
（1）． 0
（2）． $2 \sqrt{5}$

## 【解析】

## 【分析】

本题主要考查平面向量的应用，题目难度较大。从引入＂基向量＂入手，简化模的表现形式，利用转化与化归思想将问题逐步简化．

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详
解
1
$\lambda_{1} \overrightarrow{A B}+\lambda_{2} \overrightarrow{B C}+\lambda_{3} \overrightarrow{C D}+\lambda_{4} \overrightarrow{D A}+\lambda_{5} \overrightarrow{A C}+\lambda_{6} \overrightarrow{B D}=\left(\lambda_{1}-\lambda_{3}+\lambda_{5}-\lambda_{6}\right) \overrightarrow{A B}+\left(\lambda_{2}-\lambda_{4}+\lambda_{5}+\lambda_{6}\right) \overrightarrow{A D}$
要使 $\left|\lambda_{1} \overrightarrow{A B}+\lambda_{2} \overrightarrow{B C}+\lambda_{3} \overrightarrow{C D}+\lambda_{4} \overrightarrow{D A}+\lambda_{5} \overrightarrow{A C}+\lambda_{6} \overrightarrow{B D}\right|$ 的最小，只需要
$\left|\lambda_{1}-\lambda_{3}+\lambda_{5}-\lambda_{6}\right|=\left|\lambda_{2}-\lambda_{4}+\lambda_{5}+\lambda_{6}\right|=0$ ，此时只需要取 $\lambda_{1}=1, \lambda_{2}=-1, \lambda_{3}=1, \lambda_{4}=1, \lambda_{5}=1, \lambda_{6}=1$
此时 $\left|\lambda_{1} \overrightarrow{A B}+\lambda_{2} \overrightarrow{B C}+\lambda_{3} \overrightarrow{C D}+\lambda_{4} \overrightarrow{D A}+\lambda_{5} \overrightarrow{A C}+\lambda_{6} \overrightarrow{B D}\right|_{\text {min }}=0$
$\left|\lambda_{1} \overrightarrow{A B}+\lambda_{2} \overrightarrow{B C}+\lambda_{3} \overrightarrow{C D}+\lambda_{4} \overrightarrow{D A}+\lambda_{5} \overrightarrow{A C}+\lambda_{6} \overrightarrow{B D}\right|^{2}=\left|\left(\lambda_{1}-\lambda_{3}+\lambda_{5}-\lambda_{6}\right) \overrightarrow{A B}+\left(\lambda_{2}-\lambda_{4}+\lambda_{5}+\lambda_{6}\right) \overrightarrow{A D}\right|^{2}$
$=\left(\lambda_{1}-\lambda_{3}+\lambda_{5}-\lambda_{6}\right)^{2}+\left(\lambda_{2}-\lambda_{4}+\lambda_{5}+\lambda_{6}\right)^{2} \leq\left(\left|\lambda_{1}\right|+\left|\lambda_{3}\right|+\left|\lambda_{5}-\lambda_{6}\right|\right)^{2}+\left(\left|\lambda_{2}\right|+\left|\lambda_{4}\right|+\left|\lambda_{5}+\lambda_{6}\right|\right)^{2}$
$=\left(2+\left|\lambda_{5}-\lambda_{6}\right|\right)^{2}+\left(2+\left|\lambda_{5}+\lambda_{6}\right|\right)^{2}=8+4\left(\left|\lambda_{5}-\lambda_{6}\right|+\left|\lambda_{5}+\lambda_{6}\right|\right)+\left(\lambda_{5}-\lambda_{6}\right)^{2}+\left(\lambda_{5}+\lambda_{6}\right)^{2}$
$=8+4 \sqrt{\left(\left|\lambda_{5}-\lambda_{6}\right|+\left|\lambda_{5}+\lambda_{6}\right|\right)^{2}}+2\left(\lambda_{5}{ }^{2}+\lambda_{6}{ }^{2}\right)=12+4 \sqrt{\left(\lambda_{5}-\lambda_{6}\right)^{2}+\left(\lambda_{5}+\lambda_{6}\right)^{2}+2\left|\lambda_{5}{ }^{2}-\lambda_{6}{ }^{2}\right|}$
$=12+4 \sqrt{2\left(\lambda_{5}{ }^{2}+\lambda_{6}{ }^{2}\right)+2\left|\lambda_{5}{ }^{2}-\lambda_{6}{ }^{2}\right|}=20$

等号成立当且仅当 $\lambda_{1},-\lambda_{3}, \lambda_{5}-\lambda_{6}$ 均非负或者均非正，并且 $\lambda_{2},-\lambda_{4}, \lambda_{5}+\lambda_{6}$ 均非负或者均非正。

比如 $\lambda_{1}=1, \lambda_{2}=1, \lambda_{3}=-1, \lambda_{4}=-1, \lambda_{5}=1, \lambda_{6}=1$
则 $\left|\lambda_{1} \overrightarrow{A B}+\lambda_{2} \overrightarrow{B C}+\lambda_{3} \overrightarrow{C D}+\lambda_{4} \overrightarrow{D A}+\lambda_{5} \overrightarrow{A C}+\lambda_{6} \overrightarrow{B D}\right|_{\text {max }}=\sqrt{20}=2 \sqrt{5}$ ．
点晴：对于此题需充分利用转化与化归思想，从＂基向量＂入手，最后求不等式最值，是一道向量和不等式的综合题。

【点睛】对于平面向量的应用问题，需充分利用转化与化归思想、数形结合思想．