Cho Δ ABC . Gọi I, J là 2 điêm thỏa vecto IA + 3 vecto IC = vecto 0 , vecto JA + 2 vecto JB + 3 vecto JC = vecto 0 . Khi đó vecto BI = k vecto BJ . Vậy k = ?

\(\begin{array}{l}\overrightarrow {IA}  + 3\overrightarrow {IC}  = \vec 0 \Leftrightarrow  - \overrightarrow {AI}  + 3(\overrightarrow {AC}  - \overrightarrow {AI} ) = \vec 0 \Leftrightarrow 4\overrightarrow {AI}  = 3\overrightarrow {AC}  \Leftrightarrow \overrightarrow {AI}  = \frac{3}{4}\overrightarrow {AC} .\\\overrightarrow {BI}  = \overrightarrow {AI}  - \overrightarrow {AB}  = \frac{3}{4}\overrightarrow {AC}  - \overrightarrow {AB}  =  - \overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AC} .\\\overrightarrow {JA}  + 2\overrightarrow {JB}  + 3\overrightarrow {JC}  = \vec 0 \Leftrightarrow \overrightarrow {BA}  - \overrightarrow {BJ}  - 2\overrightarrow {BJ}  + 3(\overrightarrow {BC}  - \overrightarrow {BJ} ) = \vec 0\\ \Leftrightarrow \overrightarrow {BA}  - \overrightarrow {BJ}  - 2\overrightarrow {BJ}  + 3\overrightarrow {BC}  - 3\overrightarrow {BJ}  = \vec 0 \Leftrightarrow 6\overrightarrow {BJ}  = \overrightarrow {BA}  + 3\overrightarrow {BC} \\ \Leftrightarrow 6\overrightarrow {BJ}  =  - \overrightarrow {AB}  + 3(\overrightarrow {AC}  - \overrightarrow {AB} ) \Leftrightarrow 6\overrightarrow {BJ}  =  - 4\overrightarrow {AB}  + 3\overrightarrow {AC}  \Leftrightarrow \overrightarrow {BJ}  =  - \frac{2}{3}\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AC} \end{array}\)

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{\overrightarrow {BI}  =  - \overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AC} }\\{\overrightarrow {BJ}  =  - \frac{2}{3}\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AC} }\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\overrightarrow {BI}  =  - \overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AC} }\\{\frac{3}{2}\overrightarrow {BJ}  =  - \overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AC} }\end{array} \Rightarrow \overrightarrow {BI}  = \frac{3}{2}\overrightarrow {BJ} } \right.} \right.\)