\(3AI=AM\Rightarrow\overrightarrow{AI}=\frac{1}{2}\overrightarrow{IM}\)
a/
\(6\overrightarrow{BI}-4\overrightarrow{BA}=2\overrightarrow{BI}+4\left(\overrightarrow{BI}+\overrightarrow{AB}\right)=2\overrightarrow{BI}+4\overrightarrow{AI}\)
\(=2\overrightarrow{BI}+2\overrightarrow{IM}=2\left(\overrightarrow{BI}+\overrightarrow{IM}\right)=2\overrightarrow{BM}=\overrightarrow{BC}\)
b/
\(\overrightarrow{IM}=\frac{2}{3}\overrightarrow{AM}=\frac{2}{3}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\right)=\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(=\frac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\frac{1}{3}\overrightarrow{AC}=\frac{2}{3}\overrightarrow{AC}+\frac{1}{3}\overrightarrow{CB}=\frac{1}{3}\overrightarrow{CB}-\frac{2}{3}\overrightarrow{CA}\)
