Answer to Question #88604 in Analytic Geometry for Megha

Question

Calculate the area of a triangle whose vertices are given by (3, −1, 2), (1, −1, 2) and (4, −2, 1)

Answer 1

For A(3,-1,2), B(1,-1,2) and C(4,-2,1) vector area of triangle is:

$$\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})$$ $$\overrightarrow{AB}=(1-3)\overrightarrow{i}+(-1+1)\overrightarrow{j}+(2-2)\overrightarrow{k}=-2\overrightarrow{i}$$ $$\overrightarrow{AC}=(4-3)\overrightarrow{i}+(-2+1)\overrightarrow{j}+(1-2)\overrightarrow{k}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}$$ $$(\overrightarrow{AB} \times \overrightarrow{AC})=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ -2 & 0 & 0 \\ 1 & -1 & -1 \end{vmatrix} =0\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}=-2\overrightarrow{j}+2\overrightarrow{k}$$

vector area of triangle ABC:

$$\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})=-\overrightarrow{j}+\overrightarrow{k}$$

area of triangle ABC:

$$area=|\frac 1 2 (\overrightarrow{AB} \times \overrightarrow{AC})|=\sqrt{(-1)^2+1^2}=\sqrt 2$$