Question

Let $\sqrt{3}\hat{i}+\hat{j},\hat{i}+\sqrt{3}\hat{j}$ and $\beta \hat{i}+(1-\beta )\hat{j}$ respectively be the position vectors of the points $A,B$ and $C$ with respect to the origin $O$. If the distance of $C$ from the bisector of the acute angle between $OA$ and $OB$ is $\frac{3}{\sqrt{2}}$, then the sum of all possible values of $\beta$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$


$OA$ and $OB$ make angles of ${30}^{\circ }$ and ${60}^{\circ }$ respectively.
Equation of angle bisector of $OA$ and $OB$ is
$\frac{y-0}{x-0}=\tan {45}^{\circ }$
$\Rightarrow y=x$

Given, distance of $C$ from angle bisector $=\frac{3}{\sqrt{2}}$
$\Rightarrow \left|\frac{\beta -(1-\beta )}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}\Rightarrow |2\beta -1|=3\Rightarrow \beta =2\text{or}-1$

$\therefore$ Sum of all the possible values of $\beta$ is $2-1=1$