题目内容
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$\frac{8+\sqrt{2}}{4}$.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.
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