Question

9．已知點(diǎn)D，E分別為△ABC邊AB，AC的中點(diǎn)，F(xiàn)為DE的中點(diǎn)．則$\overrightarrow{BF}$=（　�。�

• A． $\frac{5}{6}\overrightarrow{BE}$+$\frac{1}{6}$$\overrightarrow{DC}$
• B． $\frac{5}{6}$$\overrightarrow{BE}$+$\frac{1}{3}$$\overrightarrow{DC}$
• C． $\frac{5}{6}$$\overrightarrow{BE}$$-\frac{1}{6}$$\overrightarrow{DC}$
• D． $\frac{5}{6}$$\overrightarrow{BE}$$-\frac{1}{3}$$\overrightarrow{DC}$

Answer 1

分析 用$\overrightarrow{BA}，\overrightarrow{BC}$表示出$\overrightarrow{BF}$，$\overrightarrow{BE}$，$\overrightarrow{DC}$，根據(jù)平面向量的基本定理求出系數(shù)．

解答 解：$\overrightarrow{DF}$=$\frac{1}{2}\overrightarrow{DE}$=$\frac{1}{4}$$\overrightarrow{BC}$，$\overrightarrow{BD}$=$\frac{1}{2}\overrightarrow{BA}$，∴$\overrightarrow{BF}$=$\overrightarrow{BD}+\overrightarrow{DF}$=$\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}$，
$\overrightarrow{BE}=\overrightarrow{BD}+\overrightarrow{DE}$=$\frac{1}{2}\overrightarrow{BA}$+$\frac{1}{2}\overrightarrow{BC}$，$\overrightarrow{DC}=\overrightarrow{DB}+\overrightarrow{BC}$=-$\frac{1}{2}\overrightarrow{BA}$+$\overrightarrow{BC}$．
設(shè)$\overrightarrow{BF}=λ\overrightarrow{BE}+μ\overrightarrow{DC}$，則$\overrightarrow{BF}$=（$\frac{λ}{2}-\frac{μ}{2}$）$\overrightarrow{BA}$+（$\frac{λ}{2}+μ$）$\overrightarrow{BC}$，
∴$\left\{\begin{array}{l}{\frac{λ}{2}-\frac{μ}{2}=\frac{1}{2}}\\{\frac{λ}{2}+μ=\frac{1}{4}}\end{array}\right.$，解得$\left\{\begin{array}{l}{λ=\frac{5}{6}}\\{μ=-\frac{1}{6}}\end{array}\right.$．
∴$\overrightarrow{BF}=\frac{5}{6}\overrightarrow{BE}-\frac{1}{6}\overrightarrow{DC}$．
故選C．

點(diǎn)評 本題考查了平面向量的線性運(yùn)算，平面向量的基本定理，屬于中檔題．