Find the area of a triangle having the points A(1,1,1), B(1,2,3) and C(2,3,1) as its vertices.

The correct answer is $\dfrac{\sqrt{21}}{2}$

Given vertices of the triangle are A(1,1,1), B(1,2,3) and C(2,3,1)

∴ $\overrightarrow{AB}$ = $0 \hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{AC}$ = $ \hat{i} + 2\hat{j} + 0\hat{k}$


$\overrightarrow{AB}$ $\times$ $\overrightarrow{AC}$ =
$\left| \begin{matrix}


\hat{i} & \phantom{-}\hat{j} & \phantom{-}\hat{k} \\


0 & \phantom{-}1 & \phantom{-}2\\


1 & \phantom{-}2 & \phantom{-}0\\


\end{matrix} \right|$ = $\hat{i}(0-4) -\hat{j}(0-2) +\hat{k}(0-1)$ = $-4\hat{i} + 2\hat{j} -\hat{k}$


$|\overrightarrow{AB}$ $\times$ $\overrightarrow{AC}|$ = $\sqrt{(-4)^2 +(2)^2 +1^2 }$ = $\sqrt{16 + 4 +1 }$ = $\sqrt{21}$


Area of $\triangle$ = $\dfrac{1}{2}$ $|\overrightarrow{AB}$ $\times$ $\overrightarrow{AC}|$ =
$\dfrac{1}{2} \sqrt{21}$ = $\dfrac{\sqrt{21}}{2}$