Show that area of the parallelogram whose diagonals are given by $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$ is $\frac{{|\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} |}}{2}$. Also, find the area of the parallelogram, whose diagonals are $2\widehat {\rm{i}} - \widehat {\rm{j}} + k$ and $\widehat {\rm{i}} + 3\widehat {\rm{j}} - \widehat {\rm{k}}$.

If $\overrightarrow {\rm{p}}$ and $\overrightarrow {\rm{q}}$

are adjacent sides of a parallelogram, then the area formed by parallelogram $= |\overrightarrow {\rm{p}} \times \overrightarrow {\rm{q}} |$

and then we shall obtained the desired result.

Let $ABCD$ be a parallelogram such that

$\overrightarrow {AB} = \vec p,\overrightarrow {AD} = \vec q \Rightarrow \overrightarrow {BC} = \vec q$

By triangle law of addition, we get
$\overrightarrow {AC} = \vec p + \vec q = \vec a$ ………(i)
Similarly, $\overrightarrow {{\rm{BD}}} = - \overrightarrow {\rm{p}} + \overrightarrow {\rm{q}} = \overrightarrow {\rm{b}}$ ………. (ii)

On adding Eqs. (i) and (ii), we get
$\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} = 2\overrightarrow {\rm{q}} \Rightarrow \overrightarrow {\rm{q}} = \frac{1}{2}(\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} )$

On subtracting Eq. (ii) from Eq. (i), we get
$\overrightarrow {\rm{a}} - \overrightarrow {\rm{b}} = 2\overrightarrow {\rm{p}} \Rightarrow \overrightarrow {\rm{p}} = \frac{1}{2}(\overrightarrow {\rm{a}} - \overrightarrow {\rm{b}} )$

Now, $\overrightarrow {\rm{p}} \times \overrightarrow {\rm{q}} = \frac{1}{4}(\overrightarrow {\rm{a}} - \overrightarrow {\rm{b}} ) \times (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} )$

$= \frac{1}{4}(\vec a \times \vec a + \vec a \times \vec b - \vec b \times \vec a - \vec b \times \vec b)$

$= \frac{1}{4}[\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} + \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} ]$

$= \frac{1}{2}(\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} )$

So, area of a parallelogram $ABCD = |\overrightarrow {\rm{p}} \times \overrightarrow {\rm{q}} | = \frac{1}{2}|\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} |$

Now, area of a parallelogram, whose diagonals are $2\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}$ and $\widehat {\rm{i}} + 3\widehat {\rm{j}} - \widehat {\rm{k}}$

$= \frac{1}{2}|(2\widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}) \times (\widehat {\rm{i}} + 3\widehat {\rm{j}} - \widehat {\rm{k}})|$

$= \frac{1}{2}\left\| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\2&{ - 1}&1\\1&3&{ - 1}\end{array}} \right\|$

$= \frac{1}{2}\left| {\left[ {\mid \widehat {\rm{i}}(1 - 3) - \widehat {\rm{j}}( - 2 - 1) + \widehat {\rm{k}}(6 + 1)} \right]} \right|$
$= \frac{1}{2}| - 2\widehat {\rm{i}} + 3\widehat {\rm{j}} + 7\widehat {\rm{k}}|$

$= \frac{1}{2}\sqrt {4 + 9 + 49}$
$= \frac{1}{2}\sqrt {62}$ sq units