If a vector $\overrightarrow {\rm{r}}$ has magnitude 14 and direction ratios 2,3 and -6 . Then, find the direction cosines and components of $\overrightarrow {\rm{r}}$, given that $\overrightarrow {\rm{r}}$ makes an acute angle with X-axis.

Here, $|\overrightarrow {\rm{r}} | = 14,$ $\overrightarrow {\rm{a}} = 2k,$ $\overrightarrow {\rm{b}} = 3k$ and $\overrightarrow {\rm{c}} = - 6k$

$\therefore$ Direction cosines

$l,m$ and $n$ are
$l = \frac{{\vec a}}{{|\vec r|}} = \frac{{2k}}{{14}} = \frac{k}{7}$

$m = \frac{{\overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{r}} |}} = \frac{{3k}}{{14}}$

and $n = \frac{{\vec c}}{{|\vec r|}} = \frac{{ - 6k}}{{14}} = \frac{{ - 3k}}{7}$

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

$\Rightarrow$ $\frac{{{k^2}}}{{49}} + \frac{{9{k^2}}}{{196}} + \frac{{9{k^2}}}{{49}} = 1$

$\Rightarrow$ $\frac{{4{k^2} + 9{k^2} + 36{k^2}}}{{196}} = 1$
$\Rightarrow$ ${k^2} = \frac{{196}}{{49}} = 4$

$\Rightarrow$ $k = \pm 2$

So, the direction cosines are $(l,m,n)$

are $\frac{2}{7},\frac{3}{7}$ and $\frac{{ - 6}}{7}$ .

[since, $\overrightarrow {\rm{r}}$ makes an acute angle with X-axis]

$\therefore$ $\overrightarrow {\rm{r}} = (l\widehat {\rm{i}} + m\widehat {\rm{j}} + n\widehat {\rm{k}})|\overrightarrow {\rm{r}} |$

$= \left( {\frac{{ + 2}}{7}\widehat {\rm{i}} + \frac{3}{7}\widehat {\rm{j}} - \frac{6}{7}\widehat {\rm{k}}} \right) \cdot 14$

$= + 4\widehat {\rm{i}} + 6\widehat {\rm{j}} - 12\widehat {\rm{k}}$