If ${\mathrm{a}}_1=\mathrm{x}\overline{\mathrm{i}}-\overline{ \mathrm{j}}+\overline{\mathrm{k}}; {\mathrm{a}}_2= \overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{k}}$ are collinear vector, then a unit vector which is parallel to $\mathrm{x}\overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{k}}$ can be

${\mathrm{a}}_1=\mathrm{x}\overline{\mathrm{i}}-\overline{ \mathrm{j}}+\overline{\mathrm{k}}; {\mathrm{a}}_2= \overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}+\mathrm{z}\overline{\mathrm{k}}$ are collinear then

$$\begin{gathered} \frac{\mathrm{x}}{1}=\frac{-1}{\mathrm{y}}=\frac{1}{\mathrm{z}}=\lambda \left(say\right) \\ required unit vector \mathrm{parallel} to given \mathrm{vector} \frac{\lambda \overline{i}-\frac{1}{\lambda }\overline{j}+\frac{1}{\lambda }\overrightarrow{k}}{\sqrt{{\lambda }^2+\frac{1}{{\lambda }^2}+\frac{1}{{\lambda }^2}}} \end{gathered}$$

for $\lambda =1$

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