Question

The vector $\overrightarrow{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\overrightarrow{b}=\hat{i}+\hat{j}$ and $\overrightarrow{c}=\hat{j}+\hat{k}$ and $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}.$ Then, which one of the following gives the possible values of $\alpha$ and $\beta$ ?

Answer 1

We have $\overrightarrow{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}$

$\overrightarrow{b}=\hat{i}+\hat{j}$

$\overrightarrow{c}=\hat{j}+\hat{k}$

As they lie in a plane

$\Rightarrow \left|\begin{array}{ccc}\alpha & 2& \beta \\ 1& 1& 0\\ 0& 1& 1\end{array}\right|=0$

$\Rightarrow \alpha (1-0)-2(1-0)=\beta (1-0)=0$

$\Rightarrow \alpha +\beta =2$ ………….$\left(1\right)$

Given that, $b=\hat{i}+\hat{j}$ & $c=\hat{j}+\hat{k}$, the equation of bisector of $\overrightarrow{b}$ & $\overrightarrow{c}$ is

$r=\lambda (b+c)$ $|\overrightarrow{b}|=\frac{\hat{i}+\hat{j}}{\sqrt{2}}=b$

$=\lambda (\frac{\hat{i}+\hat{j}}{\sqrt{2}}+\frac{\hat{j}+\hat{k}}{\sqrt{2}})$ $|\overrightarrow{c}|=\frac{\hat{j}+\hat{k}}{\sqrt{2}}=c$

$r=\frac{\lambda }{\sqrt{2}}(\hat{i}+2\hat{j}+\hat{k})$ ………..$\left(2\right)$

Since the vector a lies in plane of b & c.

$a=b+\mu c$

$a=r=b+\mu c$

$\frac{\lambda }{\sqrt{2}}(\hat{i}+2\hat{j}+\hat{k})=(\hat{i}+\hat{j})+\mu (\hat{j}+\hat{k})$

On equating the coefficient of i both sides, we get

$\frac{\lambda }{\sqrt{2}}=1$

$\Rightarrow \lambda =\sqrt{2}$

On putting $\lambda =\sqrt{2}$ in equation $\left(2\right)$, we get

$r=\hat{i}+2\hat{j}+\hat{k}$

Since, the given vector a represents the same bisection equation r.

$\alpha =1$ & $\beta =1$

Also satisfy the equation $\left(1\right)$.