The vectors $\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 2$and $\overrightarrow {\rm{b}} = - \widehat {\rm{i}} - 2\widehat {\rm{k}}$ are the adjacent sides of

a parallelogram. The angle between its diagonals is......

We have, $\overrightarrow {\rm{a}} = 3\widehat {\rm{i}} - 2\widehat {\rm{j}} + 2\widehat {\rm{k}}$ and $\overrightarrow {\rm{b}} = - \widehat {\rm{i}} - 2\widehat {\rm{k}}$

$\therefore$ $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} = 2\widehat {\rm{i}} - 2\widehat {\rm{j}}$

and $\overrightarrow {\rm{a}} - \overrightarrow {\rm{b}} = 4\widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$

Now, let $\theta$ is the acute angle between the diagonals $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{a}} - \overrightarrow {\rm{b}}$.

$\therefore$
$= \frac{{(2\widehat {\rm{i}} - 2\widehat {\rm{j}}) \cdot (4\widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}})}}{{\sqrt 8 \sqrt {16 + 4 + 16} }}$

$= \frac{{8 + 4}}{{2\sqrt 2 \cdot 6}} = \frac{1}{{\sqrt 2 }}$

$\therefore \theta = \frac{\pi }{4}$