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. $\frac{1}{\sqrt{5}}(-\hat{j}-2\hat{k})$
• B. $(\hat{j}-2\hat{k})$
• C. $\frac{1}{\sqrt{5}}(\hat{j}+4\hat{k})$
• D. $(\hat{j}+4\hat{k})$

Answer 1

The correct option is A $\frac{1}{\sqrt{5}}(-\hat{j}-2\hat{k})$

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
$\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 3& -2& 1\end{array}\right]$
$\Rightarrow (2-2)\hat{i}-(1-(-3\left)\right)\hat{j}+(-2-6)\hat{k}$
$\Rightarrow -4\hat{j}-8\hat{k}$
Also, $|\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{(-4{)}^2+(-8{)}^2}=\sqrt{80}=4\sqrt{5}$
Hence, $\hat{n}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}=\frac{-4\hat{j}-8\hat{k}}{4\sqrt{5}}=\frac{1}{\sqrt{5}}(-\hat{j}-2\hat{k})$