Question

A unit vector perpendicular to the plane of $\overrightarrow{a}=2\hat{i}-6\hat{j}-3\hat{k}$, $\overrightarrow{b}=4\hat{i}+3\hat{j}-\hat{k}$ is

• A. $\frac{4\hat{i}+3\hat{j}-\hat{k}}{\sqrt{26}}$
• B. $\frac{2\hat{i}-6\hat{j}-3\hat{k}}{7}$
• C. $\frac{3\hat{i}-2\hat{j}+6\hat{k}}{7}$
• D. $\frac{2\hat{i}-3\hat{j}-6\hat{k}}{7}$

Answer 1

The correct option is C $\frac{3\hat{i}-2\hat{j}+6\hat{k}}{7}$

A unit vector perpendicular to the planes $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\frac{\overrightarrow{a}\times \overrightarrow{b}}{|\overrightarrow{a}\times \overrightarrow{b}|}$

Now $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -6& -3\\ 4& 3& -1\end{array}\right|=(6+9)\hat{i}-(-2+12)\hat{j}-(6+24)\hat{k}$

$=\left(15\right)\hat{i}-\left(10\right)\hat{j}-\left(30\right)\hat{k}$ and $|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{{15}^2-{(-10)}^2+{30}^2}=35$

So required unit vector is $\frac{\left(15\right)\hat{i}-10\hat{j}+30\hat{k}}{35}=\frac{3\hat{i}-2\hat{j}+6\hat{k}}{7}$