Question

2．已知△ABC是边长为2的等边三角形，点D、E分别是边AB、BC的中点，点F为DE中点，则$\overrightarrow{AF}•\overrightarrow{BC}$=$-\frac{1}{2}$．

Answer 1

分析 根据条件可得到$\overrightarrow{AD}=-\frac{1}{2}\overrightarrow{BA}$，$\overrightarrow{DF}=\frac{1}{4}（\overrightarrow{BC}-\overrightarrow{BA}）$，从而得到$\overrightarrow{AF}=-\frac{3}{4}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}$，代入$\overrightarrow{AF}•\overrightarrow{BC}$进行数量积的运算即可．

解答 解：如图，

据条件：
$\overrightarrow{AF}=\overrightarrow{AD}+\overrightarrow{DF}$
=$\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{DE}$
=$-\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{AC}$
=$-\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}（\overrightarrow{BC}-\overrightarrow{BA}）$
=$-\frac{3}{4}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}$；
∴$\overrightarrow{AF}•\overrightarrow{BC}=（-\frac{3}{4}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}）•\overrightarrow{BC}$
=$-\frac{3}{4}\overrightarrow{BA}•\overrightarrow{BC}+\frac{1}{4}{\overrightarrow{BC}}^{2}$
=$-\frac{3}{4}×2×2×\frac{1}{2}+\frac{1}{4}×4$
=$-\frac{1}{2}$．
故答案为：$-\frac{1}{2}$．

点评 考查向量加法、减法和数乘的几何意义，向量的数乘和数量积运算，以及数量积的计算公式．