Câu 1:
\(\begin{array}{l}a)\overrightarrow {AB} - \overrightarrow {AC} = \overrightarrow {CB} \\b)\overrightarrow {AB} + \overrightarrow {AC} = 2\overrightarrow {AI} \end{array}\)
Câu 2:
\(\overrightarrow {BC} + \overrightarrow {MA} = \overrightarrow 0 \Leftrightarrow \overrightarrow {MA} = - \overrightarrow {BC} \)
\(\overrightarrow {AB} - \overrightarrow {NA} - 3\overrightarrow {AC} = \overrightarrow 0 \Leftrightarrow \overrightarrow {NA} = \overrightarrow {AB} - 3\overrightarrow {AC} \)
\(\begin{array}{l} \Rightarrow \overrightarrow {MN} = \overrightarrow {MA} + \overrightarrow {AN} = \overrightarrow {MA} - \overrightarrow {NA} \\ = - \overrightarrow {BC} - \left( {\overrightarrow {AB} - 3\overrightarrow {AC} } \right) = - \overrightarrow {BC} - \overrightarrow {AB} + 3\overrightarrow {AC} \\ = \overrightarrow {CB} + \overrightarrow {BA} + 3\overrightarrow {AC} = \overrightarrow {CA} + 3\overrightarrow {AC} = - \overrightarrow {AC} + 3\overrightarrow {AC} = 2\overrightarrow {AC} \\ \Rightarrow \overrightarrow {MN} = 2\overrightarrow {AC} \end{array}\)
Vậy \(MN//AC\).
Câu 3:
\(\begin{array}{l}\overrightarrow {AD} + \overrightarrow {MB} + \overrightarrow {NA} = \overrightarrow {AD} + \overrightarrow {MA} + \overrightarrow {AB} + \overrightarrow {NB} + \overrightarrow {BA} \\ = \overrightarrow {AD} + \overrightarrow {MA} + \overrightarrow {NB} = \overrightarrow {AD} + \dfrac{1}{2}\overrightarrow {DA} + \dfrac{1}{2}\overrightarrow {CB} \\ = \overrightarrow {AD} + \dfrac{1}{2}\left( {\overrightarrow {DA} + \overrightarrow {CB} } \right) = \overrightarrow {AD} + \dfrac{1}{2}.2\overrightarrow {DA} = \overrightarrow {AD} + \overrightarrow {DA} = \overrightarrow 0 \end{array}\)
