The number of unit vectors perpendicular to the

vectors$\overrightarrow{a}=\widehat{i}+\widehat{j}\text{ and }\overrightarrow{b}=\widehat{j}+\widehat{k}\text{ is }$

: The vector perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$ is along 

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{l}\widehat{i}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\widehat{j}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\widehat{k}\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|=\widehat{i}-\widehat{j}+\widehat{k}$

$\therefore$ Unit vector is $\frac{\widehat{i}-\widehat{j}+\widehat{k}}{|\widehat{i}-\widehat{j}+\widehat{k}|}=\pm \frac{(\widehat{i}-\widehat{j}+\widehat{k})}{\sqrt{3}}.$