Question

Determine unit vector which is perpendicular to both vectors $2\hat{i}-3\hat{j}+3\hat{k}and\hat{i}+\hat{j}+\hat{k}$

• A. $\frac{5}{3}(\hat{i}-2\hat{j}+\hat{k})$
• B. $\frac{1}{3}(\hat{i}+2\hat{j}-\hat{k})$
• C. $\left(\frac{-6\hat{i}+\hat{j}+5\hat{k}}{\sqrt{6}2}\right)$
• D. $1\left(\frac{5\hat{i}-8\hat{j}+5\hat{k}}{3}\right)$

Answer 1

The correct option is C $\left(\frac{-6\hat{i}+\hat{j}+5\hat{k}}{\sqrt{6}2}\right)$

Given that,

$\overrightarrow{A}=2\hat{i}-3\hat{j}+3\hat{k}$

$\overrightarrow{B}=\hat{i}+\hat{j}+\hat{k}$

Now, for two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ if $\overrightarrow{C}$ is the vector yields a vector which is perpendicular o both vectors

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

$\overrightarrow{C}=(2\hat{i}-3\hat{j}+3\hat{k})\times (\hat{i}+\hat{j}+\hat{k})$

$\overrightarrow{C}=-6\hat{i}+\hat{j}+5\hat{k}$

Now, the unit vector in the direction of $\overrightarrow{C}$ is $\frac{\overrightarrow{C}}{|\overrightarrow{C}|}$

$|\overrightarrow{C}|=\sqrt{36+1+25}$

$|\overrightarrow{C}|=\sqrt{62}$

Now, the unit vector is

$=\frac{(-6\hat{i}+\hat{j}+5\hat{k})}{\sqrt{62}}$

Hence, this is the required solution