Cho tam giác ABC. Gọi I và J là hai điểm thoả mãn \(\overrightarrow{IA}=2\overrightarrow{IB};3\overrightarrow{JA}=2\overrightarrow{JC}=\overrightarrow{0}\).
Hãy biểu diễn \(\overrightarrow{IJ}\) theo các vecto \(\overrightarrow{b}=\overrightarrow{AB},\) \(\overrightarrow{c}=\overrightarrow{AC}\).
\(\overrightarrow{IJ}=\dfrac{2}{5}\overrightarrow{b}-2\overrightarrow{c}\).\(\overrightarrow{IJ}=\dfrac{2}{5}\overrightarrow{c}-2\overrightarrow{b}\).\(\overrightarrow{IJ}=\dfrac{5}{2}\overrightarrow{c}-2\overrightarrow{b}\).\(\overrightarrow{IJ}=\dfrac{5}{2}\overrightarrow{b}-2\overrightarrow{c}\).Hướng dẫn giải:\(\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}\)
Từ \(\overrightarrow{IA}=2\overrightarrow{IB}\Rightarrow\overrightarrow{IA}=2\left(\overrightarrow{IA}+\overrightarrow{AB}\right)\Rightarrow\overrightarrow{AI}=2\overrightarrow{AB}=2\overrightarrow{b}\)
Từ \(3\overrightarrow{JA}+2\overrightarrow{JC}=\overrightarrow{0}\Rightarrow3\overrightarrow{JA}+2\left(\overrightarrow{JA}+\overrightarrow{AC}\right)=\overrightarrow{0}\Rightarrow\overrightarrow{AJ}=\dfrac{2}{5}\overrightarrow{AC}=\dfrac{2}{5}\overrightarrow{c}\)
\(\Rightarrow\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{5}\overrightarrow{c}-2\overrightarrow{b}\)
