Question

Calculate the area of the triangle through the points (5,3,2),(-2,7,-1) and (4,-2,6)

Answer 1

Solution.

We can use Heron formula which states

The area of a triangle with sides a,b,c is equal to

$$S=\sqrt{p(p-a)(p-b)(p-c)}$$

 where $p=\frac{a+b+c}{2}.$

Using the formula to find the distance between two points $A(x_1,x_2,x_3), B(y_1,y_2,y_3)$

which is

$$AB=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}$$

we can calculate the length of sides between the three points given

let say A(5,3,2), B(-2,7,-1) and C(4,-2,6).

After that, we substitute to Heron formula.

$AB=\sqrt{(5+2)^2+(3-7)^2+(2+1)^2}=\sqrt{49+16+9}= \sqrt{74}.\newline BC=\sqrt{(-2-4)^2+(7+2)^2+(-1-6)^2}=\sqrt{36+81+49}= \sqrt{166}.\newline AC=\sqrt{(5-4)^2+(3+2)^2+(2-6)^2}=\sqrt{1+25+16}= \sqrt{42}.\newline p = \frac{ a + b + c}{ 2} = \frac{ 1}{ 2} ( 8.6 + 12.88 + 6.48 ) = 13.98\newline S=\sqrt{(13.98)(13.98 - 8.6)(13.98 - 12.88)(13.98 - 6.48) }=24.9 \text{square units} \newline \text{Answer. 24.9 square units.}$