Question

The vector$\overrightarrow{V}$ directed along the internal bisector of the angle between the vectors
$\overrightarrow{A}=3\hat{i}-4\hat{j},\overrightarrow{B}=4\hat{i}+2\hat{j}-4\hat{k}$ and $\left(\overrightarrow{V}\right|=2$ is

• A. $\frac{38\hat{i}-14\hat{j}-20\hat{k}}{\sqrt{510}}$
• B. $\frac{7}{2}\hat{i}-\hat{j}-2\hat{k}$
• C. $\frac{14\hat{i}-4\hat{j}-8\hat{k}}{\sqrt{69}}$

Answer 1

The correct option is A $\frac{38\hat{i}-14\hat{j}-20\hat{k}}{\sqrt{510}}$

The direction of internal angle bisector is given by the sum of unit vectors of both the given vectors
$\frac{\overrightarrow{A}}{\left(}=\frac{3\hat{i}-4\hat{j}}{\sqrt{3^2+(-4{)}^2}}=\frac{3}{5}\hat{i}-\frac{4}{5}\hat{j}$
similarly for B
$\frac{\overrightarrow{B}}{\left(}=\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}-\frac{2}{3}\hat{k}$
Direction of vector V=$\frac{\overrightarrow{A}}{\left(}+\frac{\overrightarrow{B}}{\left(}$
$=\frac{3}{5}\hat{i}-\frac{4}{5}\hat{j}+\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}-\frac{2}{3}\hat{k}$
$=\frac{19\hat{i}-7\hat{j}-10\hat{k}}{15}$
To get the vector V with magnitude 2,first we need to find the unit vector,which is given by
$\frac{\frac{19\hat{i}-7\hat{j}-10\hat{k}}{15}}{\sqrt{\frac{{19}^2+(-7{)}^2+(-10{)}^2}{{15}^2}}}$
$=\frac{19\hat{i}-7\hat{j}-10\hat{k}}{\sqrt{510}}$
This is a unit vector,multipling with magnitude 2 we get vector

$\overrightarrow{V}=\frac{38\hat{i}-14\hat{j}-20\hat{k}}{\sqrt{510}}$