Question

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

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

Answer 1

Answer: (D)

$\alpha =1,\beta =1$

According to the given information we can write,

$\Rightarrow \overrightarrow{a}=\lambda \times (\overrightarrow{b}+\overrightarrow{c})$

$\Rightarrow \alpha \hat{i}+2\hat{j}+\beta \hat{k}=\lambda \times (\frac{\hat{i}+\hat{j}}{\sqrt{2}}+\frac{\hat{j}+\hat{k}}{\sqrt{2}})$

$\Rightarrow \alpha \hat{i}+2\hat{j}+\beta \hat{k}=\lambda \times \left(\frac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{2}}\right)$

On comparing we get

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

Thereore, $\Rightarrow \alpha =\beta =1$