Question

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

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

Answer 1

The correct option is C $\frac{1}{3}(a+b)$

Take $A$ as origin.

Then position vectors of $A,B,C$ are given $0,a$ and $b$ respectively.

Then the centroid $G$ has position vector

$\frac{0+a+b}{3}=\frac{a+b}{3}$

$\therefore AG=\frac{a+b}{3}$