Calculate the area of the face and determine a unit normal vector for the face

Question

Three vertices defined by the coordinates $V_1 = (3,-1,3)$, $V_2 = (0,3,3)$, $V_3 = (1,-2,2)$ define a face in $\mathbb{R}^3$.

How to calculate the area of the face and determine the unit normal vector of the face?

Answer 1

You have a triangle with vertices $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_3$. All you need to do is calculate $$\begin{cases} \vec{a} = \vec{v}_2 - \vec{v}_1 \\ \vec{b} = \vec{v}_3 - \vec{v}_1 \end{cases}$$ and apply the cross product.

The cross product $\vec{a} \times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, and therefore perpendicular to the triangle plane, which means it is parallel to the triangle normal. Thus, the unit normal is $$\hat{n} = \frac{\vec{a} \times \vec{b}}{\lVert \vec{a} \times \vec{b} \rVert}$$

The magnitude of the cross product, $\lVert \vec{a} \times \vec{b} \rVert$, is equal to the area of the parallelogram (with vertices at $\vec{v}_1$, $\vec{v}_1 + \vec{a}$, $\vec{v}_1 + \vec{a} + \vec{b}$, and $\vec{v}_1 + \vec{b}$), which means it is twice the area of the triangle. Thus, the area $A$ of the triangle is $$A = \frac{\lVert \vec{a} \times \vec{b} \rVert}{2}$$

In summary: $$\begin{cases} \vec{c} = \left ( \vec{v}_2 - \vec{v}_1 \right ) \times \left ( \vec{v}_3 - \vec{v}_1 \right ) \\ A = \frac{\lVert \vec{c} \rVert}{2} \\ \hat{n} = \frac{2 \vec{c}}{A} \end{cases}$$ where $A$ is the area of the triangle, $\hat{n}$ is its unit normal vector, and $\vec{c}$ is the normal vector (whose magnitude is twice the area of the triangle).

Answer 2

For the area, you can determine the length of the 3 sides and use the formula $$A=\sqrt{p(p−a)(p−b)(p−c)}$$ where $p$ is half of the perimeter of the triangle.

For the unit normal, you can find 2 displacement vectors between 2 pairs of these points, and then calculate the cross product to get a normal vector.