Question

8．在三角形ABC中，∠B=45°，AB=2，BC=3，點(diǎn)D，F(xiàn)為AB，AC的中點(diǎn)，點(diǎn)E在BC上，且BE=2EC，則$\overrightarrow{DE}$•$\overrightarrow{BF}$的值為$\frac{8+\sqrt{2}}{4}$．

Answer 1

分析 如圖所示，$\overrightarrow{DE}=\overrightarrow{DB}+\overrightarrow{BE}$=$-\frac{1}{2}\overrightarrow{BA}+\frac{2}{3}\overrightarrow{BC}$，$\overrightarrow{BF}=\frac{1}{2}（\overrightarrow{BA}+\overrightarrow{BC}）$，$\overrightarrow{BA}•\overrightarrow{BC}$=$|\overrightarrow{BA}||\overrightarrow{BC}|$cos45°，代入計(jì)算即可得出．

解答 解：如圖所示，
∵$\overrightarrow{DE}=\overrightarrow{DB}+\overrightarrow{BE}$=$-\frac{1}{2}\overrightarrow{BA}+\frac{2}{3}\overrightarrow{BC}$，$\overrightarrow{BF}=\frac{1}{2}（\overrightarrow{BA}+\overrightarrow{BC}）$，
$\overrightarrow{BA}•\overrightarrow{BC}$=$|\overrightarrow{BA}||\overrightarrow{BC}|$cos45°=$2×3×\frac{\sqrt{2}}{2}$=3$\sqrt{2}$．
∴$\overrightarrow{DE}$•$\overrightarrow{BF}$=$（-\frac{1}{2}\overrightarrow{BA}+\frac{2}{3}\overrightarrow{BC}）$•$\frac{1}{2}（\overrightarrow{BA}+\overrightarrow{BC}）$
=$-\frac{1}{4}$${\overrightarrow{BA}}^{2}$+$\frac{1}{3}$${\overrightarrow{BC}}^{2}$+$\frac{1}{12}\overrightarrow{BA}•\overrightarrow{BC}$
=$-\frac{1}{4}$×2²+$\frac{1}{3}×{3}^{2}$+$\frac{1}{12}×3\sqrt{2}$
=2+$\frac{\sqrt{2}}{4}$．
故答案為：$\frac{8+\sqrt{2}}{4}$．

點(diǎn)評(píng) 本題考查了向量的三角形法則、平行四邊形法則、數(shù)量積運(yùn)算性質(zhì)，考查了推理能力與計(jì)算能力，屬于中檔題．