Question

Determine the area of a triangle with
vertices A (1,1,2), B (2, 4, 6), and C (0, 5, 5​

Answer 1

The area of the triangle is 6.06 square units.

Step-by-step explanation:

The vertices of the triangle are:

A (1, 1, 2); B (2, 4, 6); C (0, 5, 5)​

Now,

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \left ( 2\hat{i} + \hat{j} + \hat{k} \right )-\left ( \hat{i} + \hat{j} + 2\hat{k} \right )=\hat{i} + 3\hat{j} +4 \hat{k}$

$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \left ( 0\hat{i} + 5\hat{j} + 5\hat{k} \right )-\left ( 2\hat{i} + 4\hat{j} + 6\hat{k} \right )=-2\hat{i} + \hat{j} -1 \hat{k}$

Now,

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 1 & 3 & 4\\ -2 & 1 & -1\end{vmatrix}$

$\therefore \overrightarrow{AB} \times \overrightarrow{AC} = -7\hat{i} -7 \hat{j} +7 \hat{k}$

Now,

$\left | \overrightarrow{AB} \times \overrightarrow{AC} \right | = \sqrt{(-7^{2})+(-7^{2})+7^{2}}$

$\left | \overrightarrow{AB} \times \overrightarrow{AC} \right | = \sqrt{49+49+49}$

$\therefore \left | \overrightarrow{AB} \times \overrightarrow{AC} \right | = 12.12$

The area of the triangle is given by the formula:

$A = 1/2 \times \left | \overrightarrow{AB} \times \overrightarrow{AC} \right |$

On substituting the values, we get,

A = 1/2 × 12.12

∴ A = 6.06 square units