Find a unit vector perpendicular to $\mathbf{A}=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ both.

As we have read, $\mathbf{C}=\mathbf{A}\times \mathbf{B}$ is a vector perpendicular to both A and B. Hence, a unit vector $\hat{\mathbf{n}}$ perpendicular to A and B can be written as

$\hat{\mathbf{n}}=\frac{\mathbf{C}}{C}=\frac{\mathbf{A}\times \mathbf{B}}{|\mathbf{A}\times \mathbf{B}|}$

Here, $\mathbf{A}\times \mathbf{B}=\left|\begin{array}{ccc}\hat{\mathbf{i}}& \hat{\mathbf{j}}& \hat{\mathbf{k}}\\ 2& 3& 1\\ 1& -1& 1\end{array}\right|$

$=\hat{\mathbf{i}}(3+1)+\hat{\mathbf{j}}(1-2)+\hat{\mathbf{k}}(-2-3)=4\hat{\mathbf{i}}-\hat{\mathbf{j}}-5\hat{\mathbf{k}}$

Further,$|A\times B|=\sqrt{(4{)}^2+(-1{)}^2+(-5{)}^2}=\sqrt{42}$

The desired unit vector is

$\hat{\mathbf{n}}=\frac{\mathbf{A}\times \mathbf{B}}{|\mathbf{A}\times \mathbf{B}|}$

$\hat{\mathrm{n}}=\frac{1}{\sqrt{42}}(4\hat{\mathrm{i}}-\hat{\mathrm{j}}-5\hat{\mathrm{k}})$