Question

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

• A. $\alpha =2,\beta =2$
• B. $\alpha =1,\beta =2$
• C. $\alpha =2,\beta =1$
• D. $\alpha =1,\beta =1$

Answer 1

The correct option is D $\alpha =1,\beta =1$

$\overrightarrow{b}.(\alpha \hat{i}+2\hat{j}+\beta \hat{k})=|\hat{b}|(\sqrt{{\alpha }^2+4}+{\beta }^2)\cos \theta$
$(\alpha +2)=\left(\sqrt{2}\right)\left(\sqrt{{\alpha }^2+4+{\beta }^2}\right)\cos \theta$------------(1)
$\overrightarrow{a}.\left(\alpha \hat{i}\right)+\alpha \hat{j}+\beta \hat{k}=|\overrightarrow{a}|\left(\sqrt{2^2+4+{\beta }^2}\right)\cos \theta$
$(\alpha +\beta )=\left(\sqrt{2}\right)\sqrt{{\alpha }^2+4+{\beta }^2}\cos \theta$--------------(2)
$\frac{\alpha +2}{2+\beta }=1$
$\alpha =\beta$------------(3)
$2\hat{i}+2\hat{j}+\beta \hat{k}=\mu (\hat{i}+\hat{j})+\lambda (\hat{j}+\hat{k})$ (collineorlity)
$\mu +\lambda =2$
$\mu =\alpha \lambda =\beta$
$\alpha +\beta =2$ --------------(4)
$\alpha =1\beta =1$