Three Dimensional Geometry — Class 12 Maths Solution

It is given that, $\overrightarrow {{\rm{AB}}} = (2 - 0)\widehat {\rm{i}} + (3 - 4)\widehat {\rm{j}} + ( - 1 - 1)\widehat {\rm{k}} = 2\widehat {\rm{i}} - \widehat {\rm{j}} - 2\widehat {\rm{k}}$

$\overrightarrow {{\rm{BC}}} = (4 - 2)\widehat {\rm{i}} + (5 - 3)\widehat {\rm{j}} + (0 + 1)\widehat {\rm{k}} = 2\widehat {\rm{i}} + 2\widehat {\rm{j}} + \widehat {\rm{k}}$

$\overrightarrow {{\rm{CD}}} = (2 - 4)\widehat {\rm{i}} + (6 - 5)\widehat {\rm{j}} + (2 - 0)\widehat {\rm{k}} = - 2\widehat {\rm{i}} + \widehat {\rm{j}} + 2\widehat {\rm{k}}$

$\overrightarrow {{\rm{DA}}} = (0 - 2)\widehat {\rm{i}} + (4 - 6)\widehat {\rm{j}} + (1 - 2)\widehat {\rm{k}} = - 2\widehat {\rm{i}} - 2\widehat {\rm{j}} - \widehat {\rm{k}}$

$\therefore$ Area of quadrilateral $ABCD = |\overrightarrow {AB} \times \overrightarrow {BC} | = \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\2&{ - 1}&{ - 2}\\2&2&1\end{array}} \right|$

$= |\widehat {\rm{i}}( - 1 + 4) - \widehat {\rm{j}}(2 + 4) + \widehat {\rm{k}}(4 + 2)|$
$= |3\widehat {\rm{i}} - 6\widehat {\rm{j}} + 6\widehat {\rm{k}}|$

$= \sqrt {9 + 36 + 36} = 9$ sq units