Question

In $\Delta ABC,\overrightarrow{AB}=\hat{i}+3\hat{j}-2\hat{k},\overrightarrow{AC}=3\hat{i}-\hat{j}-2\hat{k}.$
If the bisector of $\angle BAC$ meets $BC$ at $D$ and $G$ is the centroid of $\Delta ABC$, then $|\overrightarrow{GD}|=$

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $2$

Answer 1

The correct option is A $1$

$|\overrightarrow{AB}|=|\overrightarrow{AC}|=\sqrt{14}$
$\Rightarrow \Delta ABC$ is isosceles.
So, the bisector of $\angle BAC$ coincide with the median through $A$.
$\overrightarrow{AD}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=2\hat{i}+\hat{j}-2\hat{k}$
$|\overrightarrow{AD}|=3$
Since, centroid divides the median in the ratio $2:1$, we have
$3|\overrightarrow{GD}|=|\overrightarrow{AD}|\Rightarrow |\overrightarrow{GD}|=1$