Question

The unit vector which, is perpendicular to the vectors $\overrightarrow{A}=\hat{i}-2\hat{j}+\hat{k}$ and $\overrightarrow{B}=3\hat{i}-2\hat{j}-\hat{k}$ is

• A. $\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$
• B. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• C. $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}-\hat{k})$
• D. $\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})$

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

We know that the unit vector perpendicular to two vectors is given as
$\hat{n}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}$
So, we have $\overrightarrow{A}\times \overrightarrow{B}$ as per determinant method as
$\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 1\\ 3& -2& -1\end{array}\right]$
$\Rightarrow (2-(-2\left)\right)\hat{i}-(-1-3)\hat{j}+(-2-(-6\left)\right)\hat{k}$
$\Rightarrow 4\hat{i}+4\hat{j}+4\hat{k}$
Also, $|\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{4^2+4^2+4^2}=4\sqrt{3}$
Hence, $\hat{n}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}=\frac{4\hat{i}+4\hat{j}+4\hat{k}}{4\sqrt{3}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$