A unit vector perpendicular to both \(\mathbf{i}+\mathbf{j}+\mathbf{k}\) and \(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}\) is

Given vectors \(\mathbf{b}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{c}=2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}\) Let the unit vector \(a=\frac{a}{|a|}\)
\(\mathbf{a} \cdot \mathbf{b}=0\)
\(\Rightarrow \quad \frac{\left(a_{1} \mathbf{i}+a_{2} \mathbf{j}+a_{3} \mathbf{k}\right)}{|\mathbf{a}|} \cdot(\mathbf{i}+\mathbf{j}+\mathbf{k})=0\)
\(\begin{array}{lc}\Rightarrow & a_{1}+a_{2}+a_{3}=0 \\ \text { and } & \mathbf{a} \cdot \mathbf{c}=0 \\ \Rightarrow & \frac{\left(a_{1} \mathbf{i}+a_{2} \mathbf{j}+a_{3} \mathbf{k}\right)}{|\mathbf{a}|} \cdot(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k})=0\end{array}\)
\(\Rightarrow \quad 2 a_{1}+a_{2}+3 a_{3}=0 \quad \ldots\) (iii)
On solving Eqs. (ii) and (iii) by cross multiplication method
\[
\begin{aligned}
\frac{a_{1}}{3-1} &=\frac{a_{2}}{2-3}=\frac{a_{3}}{1-2} \\
\Rightarrow \quad \frac{a_{1}}{2} &=\frac{a_{2}}{-1}=\frac{a_{3}}{-1}
\end{aligned}
\]
From Eq. (i), we get
\[
a=\frac{2 i-j-k}{\sqrt{4+1+1}}=\frac{(2 i-j-k)}{\sqrt{6}}
\]