Question

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

(a) $\hat{i}-\hat{j}+\hat{k}$

(b) $\hat{i}+\hat{j}+\hat{k}$

(c) $\frac{1}{\sqrt{3}}\left(\hat{i}+\hat{j}+\hat{k}\right)$

(d) $\frac{1}{\sqrt{3}}\left(\hat{i}-\hat{j}+\hat{k}\right)$

Answer 1

(d) $\frac{1}{\sqrt{3}}\left(\hat{i}-\hat{j}+\hat{k}\right)$

$$\begin{gathered} \text{Let:} \\ \overrightarrow{a}=\hat{i}+\hat{j}+0\hat{k} \\ \overrightarrow{b}=0\hat{i}+\hat{j}+\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& 1\end{array}\right| \\ =\hat{i}-\hat{j}+\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{1+1+1} \\ =\sqrt{3} \\ \text{Unit vector perpendicular to}\overrightarrow{a}\text{and}\overrightarrow{b}\text{=}\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}} \end{gathered}$$


Disclaimer: The answer given for this question in the textbook is incorrect.