Question

Unit vector perpendicular to vector $\overrightarrow{A}=-3\hat{i}-2\hat{j}-3\hat{k}$ and $\overrightarrow{B}=2\hat{i}+4\hat{j}+6\hat{k}$ both is-

• A. $\frac{3\hat{j}-2\hat{k}}{\sqrt{13}}$
• B. $\frac{3\hat{k}-2\hat{j}}{\sqrt{13}}$
• C. $\frac{\hat{j}+2\hat{k}}{\sqrt{13}}$
• D. $\frac{\hat{i}+3\hat{j}-2\hat{k}}{\sqrt{13}}$

Answer 1

Answer: (A)

$\overrightarrow{A}\times \overrightarrow{B}$ lies perpendicular to the plane having $\overrightarrow{A}$ and $\overrightarrow{B}$ .
Thus,
$\overrightarrow{A}\times \overrightarrow{B}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& -2& -3\\ 2& 4& 6\end{array}\right]$
$\Rightarrow \overrightarrow{A}\times \overrightarrow{B}=0\hat{i}+12\hat{j}-8\hat{k}$
$\Rightarrow \overrightarrow{A}\times \overrightarrow{B}=12\hat{j}-8\hat{k}$ .
The unit vector of $\overrightarrow{A}\times \overrightarrow{B}$ is
$\Rightarrow \frac{\overrightarrow{A}\times \overrightarrow{B}}{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}=\frac{12\hat{j}-8\hat{k}}{\sqrt{208}}$
$\Rightarrow \frac{\overrightarrow{A}\times \overrightarrow{B}}{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}=\frac{12\hat{j}-8\hat{k}}{4\sqrt{13}}$
$\Rightarrow \frac{\overrightarrow{A}\times \overrightarrow{B}}{\left|\overrightarrow{A}\times \overrightarrow{B}\right|}=\frac{3\hat{j}-2\hat{k}}{\sqrt{13}}$
Therefore, the unit vector perpendicular to vector $\overrightarrow{A}=-3\hat{i}-2\hat{j}-3\hat{k}$ and $\overrightarrow{B}=2\hat{i}+4\hat{j}+6\hat{k}$ both is $\frac{3\hat{j}-2\hat{k}}{\sqrt{13}}$ .