Bài 1: b) \(\begin{array}{l}
\overrightarrow {MN} + 2\overrightarrow {PO} + \overrightarrow {MQ} = \overrightarrow 0 \\
\Leftrightarrow \overrightarrow {MP} + 2\overrightarrow {PO} = \overrightarrow 0
\end{array}\) (do MNPQ là hình bình hành)
\(\begin{array}{l}
\Leftrightarrow 2\overrightarrow {MO} + 2\overrightarrow {PO} = \overrightarrow 0 \\
\Leftrightarrow 2\overrightarrow {MO} - 2\overrightarrow {MO} = \overrightarrow 0 \\
\Leftrightarrow \overrightarrow 0 = \overrightarrow 0
\end{array}\) (luôn đúng)
Bài 2: a) Ta có:
\(\begin{array}{l}
\overrightarrow {{\rm{IJ}}} = \overrightarrow {IA} + \overrightarrow {{\rm{AJ}}} = - \overrightarrow {AI} + \overrightarrow {{\rm{AJ}}} = - \frac{1}{3}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \\
\overrightarrow {JK} = \overrightarrow {JC} + \overrightarrow {CK} = \frac{1}{2}\overrightarrow {AC} + \overrightarrow {BC} = \frac{1}{2}\overrightarrow {AC} + \overrightarrow {AC} - \overrightarrow {AB} = \frac{3}{2}\overrightarrow {AC} - \overrightarrow {AB}
\end{array}\)
b) Cần chứng minh: \(\overrightarrow {IK} = k\overrightarrow {{\rm{IJ}}} \)
Ta có: \(\overrightarrow {IK} = \overrightarrow {IB} + \overrightarrow {BK} = \frac{2}{3}\overrightarrow {AB} + 2(\overrightarrow {BA} + \overrightarrow {AC} ) = - \frac{4}{3}\overrightarrow {AB} + 2\overrightarrow {AC} = 4( - \frac{1}{3}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} ) = 4\overrightarrow {{\rm{IJ}}} \)
\( \Rightarrow \) 3 điểm I, J, K thẳng hàng.
