Question

Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

Answer 1

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are the position vectors of the vertices $A,B$ and $C$ respectively.
Then we know that the position vector of the centroid $O$ of the triangle is $\frac{\overrightarrow{a}+\overrightarrow{b}\hspace{0.17em}+\overrightarrow{c}}{3}$.
Therefore sum of the three vectors $\overrightarrow{OA},\overrightarrow{OB}$ and $\overrightarrow{OC}$ is
$$\begin{gathered} \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{a}-\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right)+\overrightarrow{b}-\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right)+\overrightarrow{c}-\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right) \\ =\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)-3\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right) \\ =\overrightarrow{0} \end{gathered}$$
Hence, Sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.