Từ giả thiết \(\overset{\rightarrow}{M A} = k \overset{\rightarrow}{M B}\), với \(k \neq 1\), ta có:
$\begin{aligned}
&\overrightarrow{O A}-\overrightarrow{O M}=k(\overrightarrow{O B}-\overrightarrow{O M}) \Leftrightarrow(1-k) \overrightarrow{O M}=\overrightarrow{O A}-\overrightarrow{O B} . \\
&\Leftrightarrow \overrightarrow{O M}=\frac{\overrightarrow{O A}-k \overrightarrow{O B}}{1-k} .\end{aligned}$

- Ta có: \(\overset{\rightarrow}{A B} = \overset{\rightarrow}{G B} - \overset{\rightarrow}{G A} = \overset{⃗}{b} - \overset{⃗}{a}\).
- Vì \(G\) là trọng tâm của tam giác \(A B C\) nên \(\overset{\rightarrow}{G A} + \overset{\rightarrow}{G B} + \overset{\rightarrow}{G C} = \overset{\rightarrow}{0} \Rightarrow \overset{\rightarrow}{G C} = - \overset{\rightarrow}{G A} - \overset{\rightarrow}{G B} = - \overset{⃗}{a} - \overset{⃗}{b}\).
- Ta có: \(\overset{\rightarrow}{B C} = \overset{\rightarrow}{B G} + \overset{\rightarrow}{G C} = - \overset{⃗}{b} + \left(\right. - \overset{⃗}{a} - \overset{⃗}{b} \left.\right) = - \overset{⃗}{a} - 2 \overset{⃗}{b}\).
- Ta có: \(\overset{\rightarrow}{C A} = \overset{\rightarrow}{G A} - \overset{\rightarrow}{G C} = \overset{⃗}{a} - \left(\right. - \overset{⃗}{a} - \overset{⃗}{b} \left.\right) = 2 \overset{⃗}{a} + \overset{⃗}{b}\).