Cho tam giác ABC có trọng tâm G.Gọi I,J lần lượt là 2 điểm thỏa mãn \(\overrightarrow{IB}\) =\(\overrightarrow{BA}\) , \(\overrightarrow{JA}\) =\(\frac{-2}{3}\) \(\overrightarrow{JC}\)
a, CMR \(\overrightarrow{IJ}\) =\(\frac{2}{5}\) \(\overrightarrow{AC}\)-\(2\overrightarrow{AB}\)
b, Tính \(\overrightarrow{IG}\) theo \(\overrightarrow{AB}\), \(\overrightarrow{AC}\)
HELP ME SẮP PHẢI NỘP RỒI
\(\overrightarrow{JA}=-\frac{2}{3}\overrightarrow{JC}\Rightarrow\overrightarrow{JA}=\frac{2}{5}\overrightarrow{CA}\)
\(\overrightarrow{IB}=\overrightarrow{BA}\Rightarrow\overrightarrow{IA}=2\overrightarrow{BA}\)
a/ \(\overrightarrow{IJ}=\overrightarrow{IA}+\overrightarrow{AJ}=2\overrightarrow{BA}-\frac{2}{5}\overrightarrow{CA}=\frac{2}{5}\overrightarrow{AC}-2\overrightarrow{AB}\)
b/Theo tính chất trọng tâm \(3\overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{AC}\Rightarrow\overrightarrow{AG}=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{IG}=\overrightarrow{IA}+\overrightarrow{AG}=2\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AC}-\frac{5}{3}\overrightarrow{AB}\)