Find the angle between the vectors $2\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}$ and $3\widehat {\rm{i}} + 4\widehat {\rm{j}} - \widehat {\rm{k}}$.

If $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$ are two vectors, making angle $\theta$ with each other,

then $\cos \theta = \frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{a}} ||\overrightarrow {\rm{b}} |}}$,

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

and $\overrightarrow {\rm{b}} = 3\widehat {\rm{i}} + 4\widehat {\rm{j}} - \widehat {\rm{k}}$

As we know. anale between two vectors $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$ is

given by
$\cos \theta = \frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{a}} ||\overrightarrow {\rm{b}} |}}$

$= \frac{{(2\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}})(3\widehat {\rm{i}} + 4\widehat {\rm{j}} - \widehat {\rm{k}})}}{{\sqrt {4 + 1 + 1} \sqrt {9 + 16 + 1} }}$

$= \frac{{6 - 4 - 1}}{{\sqrt 6 \sqrt {26} }} = \frac{1}{{2\sqrt {39} }}$

$\therefore$ $\theta = {\cos ^{ - 1}}\left( {\frac{1}{{2\sqrt {39} }}} \right)$