Question

Let G be the centroid of a-triangle ABC. If $\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{AC}=\overrightarrow{b}$ , then the bisector $\overrightarrow{AB}$ in terms of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is

• A. $\frac{2}{3}(\overrightarrow{a}+\overrightarrow{b})$
• B. $\frac{1}{6}(\overrightarrow{a}+\overrightarrow{b})$
• C. $\frac{1}{3}(\overrightarrow{a}+\overrightarrow{b})$
• D. $\frac{1}{2}(\overrightarrow{a}+\overrightarrow{b})$

Answer 1

Answer: (C)

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

Answer 2

Take $A$ as origin. Then position vectors of $A$, $B$, $C$ are $\vec{0}$, $\vec{a}$, $\vec{b}$ respectively. $\therefore$ position vector of centroid is $\frac{0+\vec{a}+\vec{b}}{3}$ i.e., $\frac{\vec{a}+\vec{b}}{3}$ $\therefore \overrightarrow{AG}=\frac{\vec{a}+\vec{b}}{3}$