Question

Find the area of the triangle with vertices $A\left( 1,1,2 \right),B\left( 2,3,5 \right)$ and $C\left( 1,5,5 \right)$.

Answer 1

Hint: We first try to find the sides of the tringle. The vector form of the sides gives the formula for the area as $\Delta ABC=\dfrac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{BC} \right|$. We use the determinant form to find the cross product of the sides. The half of the modulus gives us the area.

Complete step by step answer:
We first convert them to vector form and find the sides AB and BC.Therefore,
\[\overrightarrow{AB}=\left( 2-1 \right)\widehat{i}+\left( 3-1 \right)\widehat{j}+\left( 5-2 \right)\widehat{k} \\
\Rightarrow \overrightarrow{AB} =\widehat{i}+2\widehat{j}+3\widehat{k}\]
And,
\[\overrightarrow{BC}=\left( 1-2 \right)\widehat{i}+\left( 5-3 \right)\widehat{j}+\left( 5-5 \right)\widehat{k} \\
\Rightarrow \overrightarrow{BC} =-\widehat{i}+2\widehat{j}\]

The area of the $\Delta ABC$ will be $\Delta ABC=\dfrac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{BC} \right|$.
$\overrightarrow{AB}\times \overrightarrow{BC}=\left| \begin{matrix}
   \widehat{i} & \widehat{j} & \widehat{k} \\
   1 & 2 & 3 \\
   -1 & 2 & 0 \\
\end{matrix} \right|=\left( 0-6 \right)\widehat{i}+\left( 0+3 \right)\widehat{j}+\left( 2+2 \right)\widehat{k}=-6\widehat{i}+3\widehat{j}+4\widehat{k}$.
Now,
$\left| \overrightarrow{AB}\times \overrightarrow{BC} \right|=\sqrt{{{\left( -6 \right)}^{2}}+{{3}^{2}}+{{4}^{2}}} \\
\Rightarrow \left| \overrightarrow{AB}\times \overrightarrow{BC} \right|=\sqrt{61}$
The area of the triangle will be,
$\Delta ABC=\dfrac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{BC} \right| \\
\therefore \Delta ABC=\dfrac{61}{2}$

Therefore, the area of the triangle with vertices $A\left( 1,1,2 \right),B\left( 2,3,5 \right)$ and $C\left( 1,5,5 \right)$ is $\dfrac{61}{2}$ square units.

Note: The main condition for the formula $\Delta ABC=\dfrac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{BC} \right|$ to work is that we have broken the sides of the into their projection forms which are in perpendicular to each other. Therefore, it is quite similar to the area form of 2-D coordinate points.