Question

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

and (4, −2, 1).

Answer 1

We must find the area of a triangle whose vertices are given by (3, −1, 2), (1, −1, 2) 

and (4, −2, 1).

A(3, -1, 2); B(1, -1, 2); C(4, -2, 1)

$S = \delta ABC = \large\frac{1}{2}$$|\overrightarrow{AB}*\overrightarrow{AC} |$

$\overrightarrow{AB} = -2\overrightarrow{i}$ $\overrightarrow{AC} =\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}$

$\overrightarrow{AB}*\overrightarrow{AC} =$ $\space\space\space \space\space\space \space\space\space \space\space\space \overrightarrow{i} \space\space\space \overrightarrow{j}\space\space\space \overrightarrow{k}$

1 -1 -1 $\to 2\overrightarrow{j}-2\overrightarrow{k}$

-2 0 0

$Area = \frac{1}{2}|2\overrightarrow{j}-2\overrightarrow{k}| = \frac{1}{2}\sqrt{(2)^2+(-2)^2}=\sqrt{2}$

Answer $\sqrt{2}$ sq. units