Question

A unit vector perpendicular to $\hat{i}-2\hat{j}+\hat{k}$ and $\widehat{3i}-\hat{j}+2\hat{k}$ is

• A. $\frac{5\hat{i}+3\hat{j}+7\hat{k}}{\sqrt{83}}$
• B. $\frac{-3\hat{i}+\hat{j}+5\hat{k}}{\sqrt{35}}$
• C. $\frac{5\hat{i}+3\hat{j}-7\hat{k}}{\sqrt{83}}$
• D. $\frac{3\hat{i}-5\hat{j}+7\hat{k}}{\sqrt{83}}$

Answer 1

The correct option is B $\frac{-3\hat{i}+\hat{j}+5\hat{k}}{\sqrt{35}}$

$\text{Step 1:}$ $\text{Unit Vector}$

For a unit vector perpendicular to two given vector is given by

$\hat{C}=\frac{(\overrightarrow{A}\times \overrightarrow{B})}{|\overrightarrow{A}\times \overrightarrow{B}|}$

Here, $\hat{C}$ is the unit vector perpendicular to $\overrightarrow{A}$ and $\overrightarrow{B}$

$\text{Step 2: Calculate vector}$ $\overrightarrow{C}$

$\overrightarrow{C}=\overrightarrow{A}\times \overrightarrow{B}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 1\\ 3& -1& 2\end{array}\right|$

$=\hat{i}(-4+1)-\hat{j}(2-3)+\hat{k}(-1+6)$

$\overrightarrow{C}=-3\hat{i}+\hat{j}+5\hat{k}$

$|\overrightarrow{C}|=\sqrt{(-3{)}^2+(1{)}^2+(5{)}^2}$$=\sqrt{35}$

$\text{Step 3: Calculate}$ $\hat{C}$

$\hat{C}=\frac{\overrightarrow{C}}{|\overrightarrow{C}|}$

$\hat{C}=\frac{-3\hat{i}+\hat{j}+5\hat{k}}{\sqrt{35}}$

Hence option B is correct.