vecto DM=veto DB+vecto BM
=-vecto BD+1/2vecto BC
=-2vecto BG+1/2vecto BC
GọiK là giao của BG với AC
=>Klà trung điểm của AC
=>\(\overrightarrow{BG}=\dfrac{2}{3}\cdot\overrightarrow{BK}\)
=>\(\overrightarrow{DM}=-2\cdot\dfrac{2}{3}\cdot\overrightarrow{BK}+\dfrac{1}{2}\overrightarrow{BC}\)
\(=\dfrac{-4}{3}\cdot\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BC}\right)+\dfrac{1}{2}\overrightarrow{BC}\)
\(=\dfrac{-2}{3}\overrightarrow{BA}-\dfrac{2}{3}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{1}{6}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{1}{6}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{5}{6}\overrightarrow{AB}-\dfrac{1}{6}\overrightarrow{AC}\)
