A unit vector perpendicular to both the vectors \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is

Let \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Now, \(\quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|\) \(=\hat{\mathbf{i}}(1-0)-\hat{\mathbf{j}}(1-0)+\hat{\mathbf{k}}(1-0)\) \(=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
and \(\quad|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=\sqrt{1^{2}+(-1)^{2}+1^{2}}=\sqrt{3}\)
\(\therefore\) Required unit vector \(=\frac{\overrightarrow{\mathbf{a} \times \overrightarrow{\mathbf{b}}}}{\mid \overrightarrow{\mathbf{a} \times \overrightarrow{\mathbf{b}} \mid}}\)
\(=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}\)