902

在平面四边形 ${\displaystyle ABCD}$ 中，${\displaystyle E,F}$ 分别为 ${\displaystyle AD,BC}$ 的中点. 若 ${\displaystyle AB=2,CD=3}$, 且 ${\displaystyle \overrightarrow{EF}\cdot \overrightarrow{AB}=4}$, 则 ${\displaystyle |\overrightarrow{EF}|=}$ _____.

由题意$$\begin{array}{r}\overrightarrow{EF}=\overrightarrow{EA}+\overrightarrow{AB}+\overrightarrow{BF},\overrightarrow{EF}=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CF},\end{array}$$
故两式相加得$$\begin{array}{r}\overrightarrow{EF}=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{DC}).\end{array}$$
两边同时点乘 ${\displaystyle \overrightarrow{AB}}$:$$\begin{array}{r}4=\frac{1}{2}(4+\overrightarrow{DC}\cdot \overrightarrow{AB})\Rightarrow \overrightarrow{DC}\cdot \overrightarrow{AB}=4.\end{array}$$
从而$$\begin{array}{r}{\overrightarrow{EF}}^2=\frac{1}{4}({\overrightarrow{AB}}^2+{\overrightarrow{DC}}^2+2\overrightarrow{AB}\cdot \overrightarrow{DC})=\frac{21}{4}\Rightarrow |\overrightarrow{EF}|=\frac{\sqrt{21}}{2}.\end{array}$$