Vector Algebra — Class 12 Maths Solution

Let $\overrightarrow {AB} = 2\widehat i - 4\widehat j + 5\widehat k\,\,{\rm{and}}\,\,\overrightarrow {BC} = \widehat i - 2\widehat j - 3\widehat k$

$\therefore$ $\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} = 3\hat i - 6\widehat j + 2\hat k$
$\therefore$

Unit vector parallel to $AC = \cfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \cfrac{1}{7}(3\hat i - 6\widehat j + 2\hat k)$

Now, $\overrightarrow {AB} \times \overrightarrow {BC} = \left| {\begin{array}{llllllllllllllllllll}{\hat i}&{\hat j}&{\hat k}\\2&{ - 4}&5\\l&{ - 2}&{ - 3}\end{array}} \right|$

$= \hat i(12 + 10) - \widehat j( - 6 - 5) + \hat k( - 4 + 4) = 22\hat i + 11\hat j$
$\therefore$

Area of parallelogram $= \left| {\overrightarrow {AB} \times \overrightarrow {BC} } \right|$

$= \sqrt {{{(22)}^2} + {{(11)}^2} + 0} \sqrt {4 + 1} \sqrt 5$ sq. units