Question

The unit vector perpendicular to $\overrightarrow{A}=2\hat{i}+3\hat{j}+\hat{k}$ and $\overrightarrow{B}=\hat{i}-\hat{j}+\hat{k}$ is:

• A. $\frac{4\hat{i}-\hat{j}-5\hat{k}}{\sqrt{42}}$
• B. $\frac{4\hat{i}-\hat{j}+5\hat{k}}{\sqrt{42}}$
• C. $\frac{4\hat{i}+\hat{j}+5\hat{k}}{\sqrt{42}}$
• D. $\frac{4\hat{i}+\hat{j}-5\hat{k}}{\sqrt{42}}$

Answer 1

Answer: (A)

$\frac{4\hat{i}-\hat{j}-5\hat{k}}{\sqrt{42}}$

$\hat{u}\perp \overrightarrow{A}$ and $\hat{u}\perp \overrightarrow{B}$

$\Rightarrow \hat{u}=\frac{\overrightarrow{A}\times \overrightarrow{B}}{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}\overrightarrow{A}=2\hat{i}+3\hat{j}+\hat{k}\overrightarrow{B}=\hat{i}-\hat{j}+\hat{k}\hat{i}\times \hat{i},\hat{j}\times \hat{j}and\hat{k}\times \hat{k}=0\hat{u}=\frac{-2\times (\hat{i}\times \hat{j})+2(\hat{i}\times \hat{k})+3(\hat{j}\times \hat{k})+3(\hat{j}\times \hat{i})+(\hat{k}\times \hat{i})-(\hat{k}\times \hat{j})}{\sqrt{2^2+3^2+1^2}\times \sqrt{1^2+{(-1)}^2+1^2}}$

$=\frac{-2\hat{k}-2\hat{j}+3\hat{i}-3\hat{k}+\hat{j}+\hat{i}}{\sqrt{14}\times \sqrt{3}}=\frac{4\hat{i}-\hat{j}-5\hat{k}}{\sqrt{42}}$