What is the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) ?

Concept:

Calculations:

Given, triangle whose vertices are at A = (3, -1, 2) = $3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}}$

B = (1, -1, -3) =  $\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}-3\overrightarrow{\mathrm{k}}$

and C = (4, -3, 1) =  $4\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}$

Let A, B , and C be the vertices of the $\mathrm{△}ABC$ , then the area of that triangle = ${\displaystyle \frac{1}{2}}\times |\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}|$

$\overrightarrow{\mathrm{A}\mathrm{B}}=(\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}-3\overrightarrow{\mathrm{k}})-(3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}})$

⇒$\overrightarrow{\mathrm{A}\mathrm{B}}=-2\overrightarrow{\mathrm{i}}+0\overrightarrow{\mathrm{k}}-5\overrightarrow{\mathrm{k}}$

$\overrightarrow{\mathrm{A}\mathrm{C}}=(4\overrightarrow{\mathrm{i}}-3\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})-(3\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+2\overrightarrow{\mathrm{k}})$

⇒ $\overrightarrow{\mathrm{A}\mathrm{C}}=\overrightarrow{\mathrm{i}}-2\overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}}$

Now, $\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}=$$\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ -2& 0& -5\\ 1& -2& -1\end{array}\right|$

⇒$\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}=\overrightarrow{\mathrm{i}}(-10)-\overrightarrow{\mathrm{j}}(2+5)+\overrightarrow{\mathrm{k}}(4-0)$

⇒$\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}=-10\overrightarrow{\mathrm{i}}-7\overrightarrow{\mathrm{j}}+4\overrightarrow{\mathrm{k}}$

⇒ $|\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}|=\sqrt{(-10{)}^2+(-7{)}^2+(4{)}^2}$

⇒$|\overrightarrow{\mathrm{A}\mathrm{B}}\times \overrightarrow{\mathrm{A}\mathrm{C}}|=\sqrt{165}$

The area of that triangle = ${\displaystyle \frac{1}{2}}\times |\overrightarrow{AB}\times \overrightarrow{AC}|$

⇒ The area of that triangle = ${\displaystyle \frac{1}{2}}\times \sqrt{165}$
Hence, the area of a triangle whose vertices are at (3, -1, 2), (1, -1, -3) and (4, -3, 1) is ${\displaystyle \frac{\sqrt{165}}{2}}$