Vector Algebra — Class 12 Maths Solution

Here,$\overrightarrow {BC} = (\hat i + 5\hat j + 5\hat k) - (2\hat i + 3\hat j + 5\hat k) = - \hat i + 2\hat j$

$\overrightarrow {BA} = (\hat i + \hat j + 2\hat k) - (2\hat i + 3\hat j + 5\hat k) = - \hat i - 2\hat j - 3\hat k$

$\therefore$ $\overrightarrow {BC} \times \overrightarrow {BA} = \left| {\begin{array}{cccccccccccccccccccc}{\hat i}&{\hat j}&{\hat k}\\{ - 1}&2&0\\{ - 1}&{ - 2}&{ - 3}\end{array}} \right|$

$= ( - 6 + 0)\hat i - (3 + 0)\hat j + (2 + 2)\hat k = - 6\hat i - 3\hat j + 4\hat k$

So, $|\overrightarrow {BC} \times \overrightarrow {BA} | = \sqrt {36 + 9 + 16} = \sqrt {61}$
$\therefore$

Area of $\Delta ABC = \cfrac{1}{2}|\overrightarrow {BC} \times \overrightarrow {BA} | = \cfrac{1}{2}(\sqrt {61} )$ sq. units