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. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$1$

$O=(0,0,0)$
$\overrightarrow{OA}=\sqrt{3}\hat{i}+\hat{j}$
$\overrightarrow{OB}=\hat{i}+\sqrt{3}\hat{j}$
$\overrightarrow{OC}=\beta \hat{i}+(1-\beta )\hat{j}$
inclination$\left(\theta \right)$ of $OA={\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}$
inclination$\left(\theta \right)$ of $OB={\tan }^{-1}\frac{\sqrt{3}}{1}=\frac{\pi }{3}$
inclination$\left(\theta \right)$ of bisector of acute angle between $OA$ and $OB$ $=\frac{1}{2}(\frac{\pi }{6}+\frac{\pi }{3})=\frac{\pi }{4}$

Slope of bisector of acute angle between $OA$ and $OB$ $=\tan \frac{\pi }{4}=1$

Equation of bisector of acute angle between $OA$ and $OB$ $\Rightarrow (y-0)=\left(\tan \frac{\pi }{4}\right)(x-0)\Rightarrow x-y=0$

Distance between $C(\beta ,1-\beta )$ and $x-y=0$ is $\frac{3}{\sqrt{2}}$
$\Rightarrow \frac{|\beta -(1-\beta )|}{\sqrt{1^2+(-1{)}^2}}=\frac{3}{\sqrt{2}}$
$\Rightarrow |2\beta -1|=3$
$\Rightarrow 2\beta -1=\pm 3$
$\Rightarrow \beta =\frac{\pm 3+1}{2}$
$\Rightarrow \beta =2,-1$
$\therefore$ sum of all possible values of $\beta =2+(-1)=1$