Khi đó −→ I J = k −→ I O , vậy k = ?
\(3\overrightarrow {IA} + 2\overrightarrow {IC} - 2\overrightarrow {ID} = \vec 0 \Leftrightarrow 3\overrightarrow {IA} + 2\left( {\overrightarrow {IC} - \overrightarrow {ID} } \right) = \vec 0\)
\( \Leftrightarrow 3\overrightarrow {IA} + 2\overrightarrow {DC} = \vec 0 \Leftrightarrow - 3\overrightarrow {AI} + 2\overrightarrow {AB} = \vec 0 \Leftrightarrow \overrightarrow {AI} = \frac{2}{3}\overrightarrow {AB} \).
\(\overrightarrow {JA} - 2\overrightarrow {JB} + 2\overrightarrow {JC} = \vec 0 \Leftrightarrow \overrightarrow {AJ} - 2\left( {\overrightarrow {JC} - \overrightarrow {JB} } \right) = \vec 0 \Leftrightarrow \overrightarrow {AJ} = 2\overrightarrow {BC} \Leftrightarrow \overrightarrow {AJ} = 2\overrightarrow {AD} \).
\(\overrightarrow {IO} = \overrightarrow {AO} - \overrightarrow {AI} = \frac{1}{2}\overrightarrow {AC} - \frac{2}{3}\overrightarrow {AB} = \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AD} } \right) - \frac{2}{3}\overrightarrow {AB} = - \frac{1}{6}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AD} \).
\(\overrightarrow {IJ} = \overrightarrow {AJ} - \overrightarrow {AI} = 2\overrightarrow {AD} - \frac{2}{3}\overrightarrow {AB} = - \frac{2}{3}\overrightarrow {AB} + 2\overrightarrow {AD} \).
Ta có: \(\left\{ {\begin{array}{*{20}{l}}\begin{array}{l}\overrightarrow {IO} = - \frac{1}{6}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AD} \\\overrightarrow {IJ} = - \frac{2}{3}\overrightarrow {AB} + 2\overrightarrow {AD} \end{array}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}6\overrightarrow {IO} = - \overrightarrow {AB} + 3\overrightarrow {AD} \\\frac{3}{2}\overrightarrow {IJ} = - \overrightarrow {AB} + 3\overrightarrow {AD} \end{array}\end{array} \Rightarrow 6\overrightarrow {IO} = \frac{3}{2}\overrightarrow {IJ} \Leftrightarrow \overrightarrow {IJ} = 4\overrightarrow {IO} } \right.} \right.\).
Đáp án: 4.