Find the unit vector in the direction of sum of vectors $\overrightarrow {\rm{a}} = 2\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}$ and $\overrightarrow {\rm{b}} = 2\widehat {\rm{j}} + \widehat {\rm{k}}$.

As we know, unit vector in the direction of a vector
$\overrightarrow {\rm{a}}$ is $\frac{{\overrightarrow {\rm{a}} }}{{\left| {\overrightarrow {\rm{a}} } \right|}}$.

So, first we will find the sum of vectors :

Let $\overrightarrow {\rm{c}}$ denote the sum of $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$.

We have, $\overrightarrow {\rm{c}} = \overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$

$= 2\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}} + 2\widehat {\rm{j}} + \widehat {\rm{k}} = 2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}$

$\therefore$

Unit vector in the direction of $\overrightarrow {\rm{c}} = \frac{{\overrightarrow {\rm{c}} }}{{|\overrightarrow {\rm{c}} |}}$

$= \frac{{2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}}}{{\sqrt {{2^2} + {1^2} + {2^2}} }} = \frac{{2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}}}{{\sqrt 9 }}$

$\widehat {\rm{c}} = \frac{{2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}}}{3}$