Question

he 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 bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$. Then, which of the following gives possible values of $\alpha$ and $\beta$?

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

Answer 1

The correct option is A

$\alpha =1,\beta =1$

Explanation for correct option

Given, 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 bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$.

Therefore,

$$\begin{gathered} \overrightarrow{a}=\lambda (b+c) \\ \Rightarrow \alpha \hat{i}+2\hat{j}+\beta \hat{k}=\lambda (b+c) \\ \Rightarrow b=\frac{(\hat{i}+\hat{j})}{\sqrt{(1^2+1^2}}=\frac{(\hat{i}+\hat{j})}{\sqrt{2}} \\ \Rightarrow c=\frac{(\hat{j}+\hat{k})}{\sqrt{(1^2+1^2}}=\frac{(\hat{j}+\hat{k})}{\sqrt{2}} \end{gathered}$$

So,

$$\begin{gathered} \alpha \hat{i}+2\hat{j}+\beta \hat{k}=\lambda \left[\frac{(\hat{i}+\hat{j})}{\sqrt{2}}+\frac{(\hat{j}+\hat{k})}{\sqrt{2}}\right] \\ =\frac{\lambda }{\sqrt{2}}\left[\hat{i}+2\hat{j}+\hat{k}\right] \end{gathered}$$

On comparing both the sides, we get;

$$\begin{gathered} \lambda =\alpha \sqrt{2} \\ \lambda =\sqrt{2} \\ \lambda =\beta \sqrt{2} \end{gathered}$$

Therefore, $\alpha =1,\beta =1$

Hence, option (A) is correct