Question

8．在三角形ABC中，∠B=45°，AB=2，BC=3，点D，F为AB，AC的中点，点E在BC上，且BE=2EC，则$\overrightarrow{DE}$•$\overrightarrow{BF}$的值为$\frac{8+\sqrt{2}}{4}$．

Answer 1

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

解答 解：如图所示，
∵$\overrightarrow{DE}=\overrightarrow{DB}+\overrightarrow{BE}$=$-\frac{1}{2}\overrightarrow{BA}+\frac{2}{3}\overrightarrow{BC}$，$\overrightarrow{BF}=\frac{1}{2}（\overrightarrow{BA}+\overrightarrow{BC}）$，
$\overrightarrow{BA}•\overrightarrow{BC}$=$|\overrightarrow{BA}||\overrightarrow{BC}|$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+$\frac{\sqrt{2}}{4}$．
故答案为：$\frac{8+\sqrt{2}}{4}$．

点评 本题考查了向量的三角形法则、平行四边形法则、数量积运算性质，考查了推理能力与计算能力，属于中档题．