Question

The unit vector perpendicular to both $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$ is

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

Answer 1

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

Let $\overrightarrow{a}=\hat{i}+\hat{j}$ and $\overrightarrow{b}=\hat{j}+\hat{k}$

Now, $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|$

$=\hat{i}\left(1\right)-\hat{j}\left(1\right)+\hat{k}\left(1\right)=\hat{i}-\hat{j}+\hat{k}$

and $|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{{\left(1\right)}^2+{(-1)}^2+1^2}=\sqrt{3}$

Unit vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$

$=\frac{(\hat{i}+\hat{j})\times (\hat{j}+\hat{k})}{|(\hat{i}+\hat{j})\times (\hat{j}+\hat{k})|}$

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