Question

The area of the triangle whose vertices are $A = (1, -1, 2), B = (2, 1, -1)$ and $C = (3, -1, 2)$ is

• A. $4 \sqrt{5}$ s units
• B. $2 \sqrt{3}$ s units
• C. $ \sqrt{13}$ s units
• D. $ \sqrt{15}$ s units

Answer 1

The Correct Option is C

Area of $\Delta\: ABC = \frac{1}{2} | \overrightarrow{BA} \times \overrightarrow{BC} |$
$\overrightarrow{BA} = - \hat{i} - 2 \hat{j} + 3 \hat{k} $
$\overrightarrow{BC} = \hat{i} - 2 \hat{j} + 3 \hat{k} $
$\therefore$ Area = $ \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1& - 2&3\\ 1& -2 & 3 \end{vmatrix}$
$ = \frac{1}{6} | 6 \, \hat{j} + 4 \hat{k} | $
$ = |3 \hat{j} + 2 \hat{k} | = \sqrt{9 + 4} = \sqrt{13}$ s units