A vector $\overrightarrow {\rm{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\overrightarrow {\rm{r}}$ is $2\sqrt 3$ units, then find the value of $\overrightarrow {\rm{r}}$.

If a vector $\overrightarrow {\rm{r}}$ is inclined at equal angles to the three axes,

then direction cosines of vector,

$\overrightarrow {\rm{r}}$

will be same and then $\overrightarrow {\rm{r}} = \overrightarrow {\rm{r}} \cdot |\overrightarrow {\rm{r}} |$.

We have, $|\overrightarrow {\rm{r}} | = 2\sqrt 3$

As, $\overrightarrow {\rm{r}}$ is equally inclined to the three axes,

$\overrightarrow {\rm{r}}$ so direction cosines of the unit vector

$\overrightarrow {\rm{r}}$ will be same. i.e., $l = m = n$.

As we know, ${l^2} + {m^2} + {n^2} = 1$

$\Rightarrow$ ${l^2} + {l^2} + {l^2} = 1$

$\Rightarrow$ ${l^2} = \frac{1}{3}$
$\Rightarrow$ $l = \pm \left( {\frac{1}{{\sqrt 3 }}} \right)$

So $\widehat {\rm{r}} = \pm \frac{1}{{\sqrt 3 }}\widehat {\rm{i}} \pm \frac{1}{{\sqrt 3 }}\widehat {\rm{j}} \pm \frac{1}{{\sqrt 3 }}\widehat {\rm{k}}$

$\therefore$ $\quad \overrightarrow {\rm{r}} = \widehat {\rm{r}}|\overrightarrow {\rm{r}} |$

$= \left[ { \pm \frac{1}{{\sqrt 3 }}\widehat {\rm{i}} \pm \frac{1}{{\sqrt 3 }}\widehat {\rm{j}} \pm \frac{1}{{\sqrt 3 }}\widehat {\rm{k}}} \right]2\sqrt 3$

$= \pm 2\widehat {\rm{i}} \pm 2\widehat {\rm{j}} \pm 2\widehat {\rm{k}} = \pm 2(\widehat {\rm{i}} + \widehat {\rm{j}} + \widehat {\rm{k}})$