Giải thích các bước giải:
\(\begin{array}{l}
a)\,\overrightarrow {AB} = \left( { - 1;5} \right),\,\overrightarrow {AC} = \left( { - 2;1} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = \left( { - 1} \right).\left( { - 2} \right) + 5.1 = 7\\
b)\,\left\{ \begin{array}{l}
{x_I} = \frac{{{x_A} + {x_B}}}{2} = \frac{5}{2}\\
{y_I} = \frac{{{y_A} + {y_B}}}{2} = \frac{3}{2}
\end{array} \right. \Rightarrow I\left( {\frac{5}{2};\frac{3}{2}} \right)\\
\left\{ \begin{array}{l}
{x_G} = \frac{{{x_A} + {x_B} + {x_C}}}{3} = 2\\
{y_G} = \frac{{{y_A} + {y_B} + {y_C}}}{2} = 1
\end{array} \right. \Rightarrow G\left( {2;1} \right)\\
c)\,M\left( {x;y} \right)\\
\Rightarrow \overrightarrow {MA} = \left( {3 - x; - 1 - y} \right)\\
\overrightarrow {AB} + 3\overrightarrow {BC} = \left( {3; - 1} \right) + 3\left( { - 1; - 4} \right) = \left( {0; - 13} \right)\\
\Rightarrow \overrightarrow {MA} = \overrightarrow {AB} + 3\overrightarrow {BC} \\
\Leftrightarrow \left\{ \begin{array}{l}
3 - x = 0\\
- 1 - y = - 13
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 12
\end{array} \right. \Rightarrow M\left( {3;12} \right)
\end{array}\)
