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a) \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=\dfrac{-1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
b) CG.CAN??
a: \(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)
Do G là trọng tâm tam giác
\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}=-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
Do I là trung điểm AG
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right)=-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AB}=\dfrac{1}{5}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)=-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{CA}-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{CA}-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}=\dfrac{4}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
Gọi M là trung điểm BC, theo tính chất trọng tâm:
\(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}\)
Mà I là trung điểm AG \(\Rightarrow\overrightarrow{IG}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AM}\Rightarrow\overrightarrow{GI}=-\dfrac{1}{3}\overrightarrow{AM}\)
Lại có: M là trung điểm BC \(\Rightarrow\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{0}\)
Nên ta có:
\(\overrightarrow{AB}+\overrightarrow{AC}+6\overrightarrow{GI}=\overrightarrow{AM}+\overrightarrow{MB}+\overrightarrow{AM}+\overrightarrow{MC}+6.\left(-\dfrac{1}{3}\right)\overrightarrow{AM}\)
\(=2\overrightarrow{AM}-2\overrightarrow{AM}=\overrightarrow{0}\) (đpcm)
