Question

The vector $\overrightarrow{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}$ 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 $\overrightarrow{c}$, then which are of the following gives possible values of $\alpha$ and $\beta$

• A. $\alpha =2,\beta =2$
• B. $\alpha =1,\beta =1$
• C. $\alpha =2,\beta =1$
• D. $\alpha =1,\beta =0$

Answer 1

Answer: (B)

$\alpha =1,\beta =1$

Since the three vectors are coplanar, the determinant formed by their components will become zero,

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

which results in $\alpha +\beta =2.$
Also since $\overline{a}$ bisects the angle between $\overline{b}$ and $\overline{c}$, the dot product of $\overline{a}$ with $\overline{b}$ and $\overline{c}$ will be equal.

$\alpha \times 1+2\times 1=2\times 1+1\times \beta$
This results in $\alpha =\beta$
$\Rightarrow$, $\alpha =\beta =1.$