浙江 2019 · 数学 q17

题面

17. 已知正方形 ABCD 的边长为 1，当每个 $\lambda_{i}(i=1,2,3,4,5,6)$ 取遍 $\pm1$ 时，

$\left|\lambda_{1}\overrightarrow{AB}+\lambda_{2}\overrightarrow{BC}+\lambda_{3}\overrightarrow{CD}+\lambda_{4}\overrightarrow{DA}+\lambda_{5}\overrightarrow{AC}+\lambda_{6}\overrightarrow{BD}\right|$ 的最小值是____；最大值是____.

17.已知正方形ABCD的边长为1，当每个 $\lambda_{i}(i = 1,2,3,4,5,6)$ 取遍 $\pm 1$ 时，$|\lambda_1\overrightarrow{AB} + \lambda_2\overrightarrow{BC} + \lambda_3\overrightarrow{CD} + \lambda_4\overrightarrow{DA} + \lambda_5\overrightarrow{AC} + \lambda_6\overrightarrow{BD}|$ 的最小值是 ____；最大值是 ____。

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.

$\lambda_{1} \overrightarrow{A B} + \lambda_{2} \overrightarrow{B C} + \lambda_{3} \overrightarrow{C D} + \lambda_{4} \overrightarrow{D A} + \lambda_{5} \overrightarrow{A C} + \lambda_{6} \overrightarrow{B D} = \left(\lambda_{1} - \lambda_{3} + \lambda_{5} - \lambda_{6}\right) \overrightarrow{A B} + \left(\lambda_{2} - \lambda_{4} + \lambda_{5} + \lambda_{6}\right) \overrightarrow{A D}$

要使 $\left|\lambda_1\overrightarrow{AB} +\lambda_2\overrightarrow{BC} +\lambda_3\overrightarrow{CD} +\lambda_4\overrightarrow{DA} +\lambda_5\overrightarrow{AC} +\lambda_6\overrightarrow{BD}\right|$ 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 $\left|\lambda_1\overrightarrow{AB} + \lambda_2\overrightarrow{BC} + \lambda_3\overrightarrow{CD} + \lambda_4\overrightarrow{DA} + \lambda_5\overrightarrow{AC} + \lambda_6\overrightarrow{BD}\right|_{\min} = 0$

$\begin{array}{l} \left| \lambda_{1} \overline{{A B}} + \lambda_{2} \overline{{B C}} + \lambda_{3} \overline{{C D}} + \lambda_{4} \overline{{D A}} + \lambda_{5} \overline{{A C}} + \lambda_{6} \overline{{B D}} \right|^{2} = \left| \left(\lambda_{1} - \lambda_{3} + \lambda_{5} - \lambda_{6}\right) \overline{{A B}} + \left(\lambda_{2} - \lambda_{4} + \lambda_{5} + \lambda_{6}\right) \overline{{A D}} \right|^{2} \\ = \left(\lambda_{1} - \lambda_{3} + \lambda_{5} - \lambda_{6}\right)^{2} + \left(\lambda_{2} - \lambda_{4} + \lambda_{5} + \lambda_{6}\right)^{2} \leq \left(\left| \lambda_{1} \right| + \left| \lambda_{3} \right| + \left| \lambda_{5} - \lambda_{6} \right|\right)^{2} + \left(\left| \lambda_{2} \right| + \left| \lambda_{4} \right| + \left| \lambda_{5} + \lambda_{6} \right|\right)^{2} \\ = \left(2 + \left| \lambda_{5} - \lambda_{6} \right|\right)^{2} + \left(2 + \left| \lambda_{5} + \lambda_{6} \right|\right)^{2} = 8 + 4 \left(\left| \lambda_{5} - \lambda_{6} \right| + \left| \lambda_{5} + \lambda_{6} \right|\right) + \left(\lambda_{5} - \lambda_{6}\right)^{2} + \left(\lambda_{5} + \lambda_{6}\right)^{2} \\ = 8 + 4 \sqrt{\left(\left| \lambda_{5} - \lambda_{6} \right| + \left| \lambda_{5} + \lambda_{6} \right|\right)^{2}} + 2 \left(\lambda_{5}^{2} + \lambda_{6}^{2}\right) = 1 2 + 4 \sqrt{\left(\lambda_{5} - \lambda_{6}\right)^{2} + \left(\lambda_{5} + \lambda_{6}\right)^{2} + 2 \left| \lambda_{5}^{2} - \lambda_{6}^{2} \right|} \\ = 1 2 + 4 \sqrt{2 \left(\lambda_{5}^{2} + \lambda_{6}^{2}\right) + 2 \left| \lambda_{5}^{2} - \lambda_{6}^{2} \right|} = 2 0 \end{array}$

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 $\lambda_{1} = 1, \lambda_{2} = 1, \lambda_{3} = -1, \lambda_{4} = -1, \lambda_{5} = 1, \lambda_{6} = 1$

$\left| \lambda_{1} \overrightarrow{A B} + \lambda_{2} \overrightarrow{B C} + \lambda_{3} \overrightarrow{C D} + \lambda_{4} \overrightarrow{D A} + \lambda_{5} \overrightarrow{A C} + \lambda_{6} \overrightarrow{B D} \right|_{\max} = \sqrt{2 0} = 2 \sqrt{5}.$

点睛：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.