Question

2．已知△ABC是邊長(zhǎng)為2的等邊三角形，點(diǎn)D、E分別是邊AB、BC的中點(diǎn)，點(diǎn)F為DE中點(diǎn)，則$\overrightarrow{AF}•\overrightarrow{BC}$=$-\frac{1}{2}$．

Answer 1

分析 根據(jù)條件可得到$\overrightarrow{AD}=-\frac{1}{2}\overrightarrow{BA}$，$\overrightarrow{DF}=\frac{1}{4}（\overrightarrow{BC}-\overrightarrow{BA}）$，從而得到$\overrightarrow{AF}=-\frac{3}{4}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}$，代入$\overrightarrow{AF}•\overrightarrow{BC}$進(jìn)行數(shù)量積的運(yùn)算即可．

解答 解：如圖，

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

點(diǎn)評(píng) 考查向量加法、減法和數(shù)乘的幾何意義，向量的數(shù)乘和數(shù)量積運(yùn)算，以及數(shù)量積的計(jì)算公式．