Vector Algebra — Class 12 Maths Solution

As we know,
Area of $\Delta {\rm{ABC}} = \frac{1}{2}|\overrightarrow {{\rm{AB}}} \times \overrightarrow {{\rm{AC}}} |$.

Here $\overrightarrow {{\rm{AB}}} = (2 - 1)\widehat {\rm{i}} + ( - 1 - 2)\widehat {\rm{j}} + (4 - 3)\widehat {\rm{k}}$

$= \widehat {\rm{i}} - 3\widehat {\rm{j}} + \widehat {\rm{k}}$

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

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

$\therefore$ $\overrightarrow {{\rm{AB}}} \times \overrightarrow {{\rm{AC}}} = \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\1&{ - 3}&1\\3&3&{ - 4}\end{array}} \right|$

$= \widehat {\rm{i}}(12 - 3) - \widehat {\rm{j}}( - 4 - 3) + \widehat {\rm{k}}(3 + 9)$

$= 9\widehat {\rm{i}} + 7\widehat {\rm{j}} + 12\widehat {\rm{k}}$

and $|\overrightarrow {{\rm{AB}}} \times \overrightarrow {{\rm{AC}}} | = \sqrt {{9^2} + {7^2} + {{12}^2}}$

$= \sqrt {81 + 49 + 144}$
$= \sqrt {274}$

$\therefore$ Area of $\Delta ABC = \frac{1}{2}|\overrightarrow {{\rm{AB}}} \times \overrightarrow {{\rm{AC}}} |$

$= \frac{1}{2}\sqrt {274} {\rm{sq}}$ units