a/ \(\overrightarrow{IC}+2\overrightarrow{IB}=\overrightarrow{IC}+2\overrightarrow{IC}+2\overrightarrow{CB}=3\overrightarrow{IC}+2\overrightarrow{CB}\)
\(3\overrightarrow{IC}+2\overrightarrow{CB}=\overrightarrow{0}\Leftrightarrow\overrightarrow{IC}=\frac{2}{3}\overrightarrow{BC}\)
Vậy lấy I sao cho \(\left\{{}\begin{matrix}\overrightarrow{IC}\uparrow\uparrow\frac{2}{3}\overrightarrow{BC}\\IC=\frac{2}{3}BC\end{matrix}\right.\)
\(\overrightarrow{v}=\overrightarrow{IC}+\overrightarrow{CA}+\overrightarrow{IC}+\overrightarrow{CB}+\overrightarrow{IC}=\frac{2}{3}.3\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{CB}\)
\(=2\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{CB}=\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}\)