Area of a triangle from vector coordinates of vertices in 3D

Question

I have three vectors $v_1, v_2$, and $v_3$, giving the vertices of a triangle. The $z$-coordinates are the same, so the $(x,y)$-coordinates alone give the vertices of an identical triangle in the $xy$-plane. Is there a general way to find the area of this triangle given the vectors?

$$ v_1 = (v_{x1}, v_{y1}, v_{z1}) \\ v_2 = (v_{x2}, v_{y2}, v_{z2}) \\ v_3 = (v_{x3}, v_{y3}, v_{z3}) $$

?

Answer 1

\begin{align*} \Delta &= \frac{1}{2}|(\mathbf{v}_{1}-\mathbf{v}_{3}) \times (\mathbf{v}_{2}-\mathbf{v}_{3})| \\ &= \frac{1}{2} |\mathbf{v}_{1} \times \mathbf{v}_{2}+ \mathbf{v}_{2} \times \mathbf{v}_{3}+ \mathbf{v}_{3} \times \mathbf{v}_{1}| \end{align*}

Answer 2

Hint: As per your assumption, if you consider the triangle to lay entirely in 2D, and the vectors represent coordinates of each vertex, then see this reference (has animation too): Area of triangle in 2D.

For a vector form, see: Triangle-Area-2d-VectorForm.

Answer 3

Its just $1/2 a\times b$ ie cross product. Where a represents vector from $v_2$ to $v_3$ and b represents vector from $v_2$ to $v_1$ so its vector area $A=1/2(v_2-v_3)\times (v_2-v_1)$ and its area in units is just the |..| of vector area.

Answer 4

If $a$ and $b$ are the vectors for two sides then you can reduce the above answers to $\dfrac{1}{2} \sqrt{|a|^2|b|^2 - (a\cdot b)^2}$ with the advantage that it works in any dimension.