Question

Let G be the centroid of ∆ ABC. If $\overrightarrow{AB}=\overrightarrow{a,}\overrightarrow{AC}=\overrightarrow{b,}$ then the bisector $\overrightarrow{AG},$ in terms of $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is
(a) $\frac{2}{3}\left(\overrightarrow{a}+\overrightarrow{b}\right)$

(b) $\frac{1}{6}\left(\overrightarrow{a}+\overrightarrow{b}\right)$

(c) $\frac{1}{3}\left(\overrightarrow{a}+\overrightarrow{b}\right)$

(d) $\frac{1}{2}\left(\overrightarrow{a}+\overrightarrow{b}\right)$

Answer 1

(c) $\frac{1}{3}\left(\overrightarrow{a}+\overrightarrow{b}\right)$

Taking $A$ as origin. Then, position vector of $A,B$ and $C$ are $\overrightarrow{0},\overrightarrow{a}$ and $\overrightarrow{b}$ respectively. Then,
Centroid $G$ has position vector $\frac{\overrightarrow{0}+\overrightarrow{a}+\overrightarrow{b}}{3}=\frac{\overrightarrow{a}+\overrightarrow{b}}{3}$
Therefore, $AG=\frac{\overrightarrow{a}+\overrightarrow{b}}{3}-\overrightarrow{0}=\frac{\overrightarrow{a}+\overrightarrow{b}}{3}$