Answer to Question #222313 in Analytic Geometry for jony

Question

Find the area of the triangle determined by the points (2,-1,0),(2,0,4) and (0,2,3)

Answer 1

Let $\overline {u}=(2-2,0-(-1),4-0)=(0,1,4)$ be the vector joining $(2,-1,0)$ and $(2,0,4).$ Let $\overline {v}=(0-2,2-(-1),3-0)=(-2,3,3)$ be the vector joining $(2,-1,0)$ and $(0,2,3).$

Let us find the vector product

$\overline{u}\times\overline{v}= \begin{vmatrix}i & j & k\\ 0 & 1 & 4\\ -2 & 3 & 3 \end{vmatrix} =-9i-8j+2k.$

It follows that the area of the triangle determined by the points$(2,-1,0),(2,0,4)$ and $(0,2,3)$ is equal to

$A=\frac{1}{2}|\overline{u}\times\overline{v}|=\frac{1}{2}\sqrt{(-9)^2+(-8)^2+2^2}=\frac{1}{2}\sqrt{149}\approx 6.1$ (square units)