Question

1．已知菱形ABCD的邊長(zhǎng)為2，∠BAD=120°，$\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}$，$\overrightarrow{DF}=\frac{1}{3}\overrightarrow{DC}$，則$\overrightarrow{AE}•\overrightarrow{AF}$=（　�。�

• A． $\frac{1}{2}$
• B． 2
• C． 1
• D． $\frac{1}{6}$

Answer 1

分析 將$\overrightarrow{AE}，\overrightarrow{AF}$分別利用菱形的邊對(duì)應(yīng)的向量表示出來，展開計(jì)算即可．

解答 解：因?yàn)榱庑蜛BCD的邊長(zhǎng)為2，∠BAD=120°，$\overrightarrow{BE}=\frac{1}{2}\overrightarrow{BC}$，$\overrightarrow{DF}=\frac{1}{3}\overrightarrow{DC}$，則$\overrightarrow{AE}•\overrightarrow{AF}$=$（\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}）（\overrightarrow{AD}+\frac{1}{3}\overrightarrow{DC}）$=$\overrightarrow{AB}•\overrightarrow{AD}$$+\frac{1}{3}\overrightarrow{AB}•\overrightarrow{DC}$$+\frac{1}{2}\overrightarrow{BC}•\overrightarrow{AD}$$+\frac{1}{6}\overrightarrow{BC}•\overrightarrow{DC}$=$2×2×（-\frac{1}{2}）+\frac{1}{3}×{2}^{2}+\frac{1}{2}×{2}^{2}+\frac{1}{6}×{2}^{2}×（-\frac{1}{2}$）=1；
故選C．

點(diǎn)評(píng) 本題考查了平面向量的三角形法則以及數(shù)量積公式的運(yùn)用；關(guān)鍵是選擇適當(dāng)?shù)鼗�底表示所求向量，進(jìn)行向量的運(yùn)算．