The vector $\overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$ lies in…
The vector $\overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$ lies in the plane of the vectors $\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}=\hat{j}+\hat{k}$ and bisects the angle between $\bar{b}$ and $\bar{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$ ?
$\alpha=1, \beta=1$
$\alpha=2, \beta=2$
$\alpha=1, \beta=2$
$\alpha=2, \beta=1$
Solution
Since $\overline{\mathrm{a}}$ bisects the angle between $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$.
$\therefore \quad$ The equation of bisector of $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ is
$\begin{aligned}
& \overline{\mathrm{a}}=\lambda(\hat{\mathrm{b}}+\hat{\mathrm{c}}) \\
& \Rightarrow \alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}=\lambda\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}+\frac{\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}}\right)
\end{aligned}$
$\Rightarrow \alpha \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}=\frac{\lambda}{\sqrt{2}}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$ On comparing, we get $\alpha=\frac{\lambda}{\sqrt{2}}, 2=\sqrt{2} \lambda$ and $\beta=\frac{\lambda}{\sqrt{2}}$ $\Rightarrow \alpha=1, \beta=1$