Question

Find the area of triangle in 3 dimensions formed by A(-1,0,1), B(-2,1,0) and C(1,1,2).

Answer 1

$\large\underline{\sf{Solution-}}$

Given vertices of $\triangle$ABC are

• Coordinates of A (- 1, 0, 1)

• Coordinates of B (- 2, 1, 0)

• Coordinates of C (1, 1, 2)

So,

$\rm \: \vec{OA} = - \hat{i} + \hat{k}$

$\rm \: \vec{OB} = - 2\hat{i} + \hat{j}$

$\rm \: \vec{OC} = \hat{i} + \hat{j} + 2\hat{k}$

Now, Consider

$\rm \: \vec{AB} = \vec{OB} - \vec{OA}$

$\rm \: \vec{AB} = - 2\hat{i} + \hat{j} + \hat{i} - \hat{k}$

$\rm\implies \:\boxed{ \rm{ \:\vec{AB} = - \hat{i} + \hat{j} - \hat{k} \: }}$

Now, Consider

$\rm \: \vec{AC} = \vec{OC} - \vec{OA}$

$\rm \: \vec{AC} = \hat{i} + \hat{j} + 2\hat{k} + \hat{i} - \hat{k}$

$\rm\implies \:\boxed{ \rm{ \:\vec{AC} = 2\hat{i} + \hat{j} + \hat{k} \: }}$

Now, Consider

$\rm \: \vec{AB} \times \vec{AC}$

$\rm \: \: \: = \: \: \begin{gathered}\sf \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\ - 1&1& - 1\\2&1&1\end{array}\right | \end{gathered}$

$\rm \: = \: \hat{i} (1 + 1) - \hat{j} ( - 1 + 2) + \hat{k} ( - 1 - 2)$

$\rm \: = \: 2\hat{i} - \hat{j} - 3\hat{k}$

So,

$\rm\implies \:\vec{AB} \times \vec{AC} = \: 2\hat{i} - \hat{j} - 3\hat{k}$

Now,

$\rm \: |\vec{AB} \times \vec{AC} |$

$\rm \: = \: \sqrt{ {(2)}^{2} + {( - 1)}^{2} + {( - 3)}^{2} }$

$\rm \: = \: \sqrt{4 + 1 + 9}$

$\rm \: = \: \sqrt{14}$

So,

$\rm\implies \: |\vec{AB} \times \vec{AC} | = \sqrt{14}$

So,

$\rm \: Area_{(\triangle\:ABC)} = \dfrac{1}{2} |\vec{AB} \times \vec{AC} |$

$\bf\implies \:Area_{(\triangle\:ABC)} = \dfrac{ \sqrt{14} }{2} \: sq. \: units$