Finding normal equation of a plane from three points

Question

I was given the following problem:

Find the normal equation of the plane $S$ that contains $P_0=(1, -1, 1), P_1=(-2, 0, 1)$, $P_2=(-1, 1, 1)$.

I was wondering whether my solution is correct. Here's what I did.

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Let $\vec{r}_0 :=<-2, 0, 1>$ be the position vector of $P_1$.

Let $\vec{v}:= <1, -1, 1> - <-1, 1, 1>=<2, -2, 0>$ be the vector representation of the line that goes from $P0$ to $P2$.

Notice that $\vec{v} \cdot \vec{w}=0 \iff 2w_1-2w_2+0w_3=0 \iff w_1=w_2$. This means any three-dimensional vector $\vec{w}$ whose first and second components are equal will be normal to $\vec{v}$ and, therefore, normal to the plane. From this we can define $\vec{n}:=[1, 1, 1]$ to be the vector normal to the plane.

All of this is enough information to provide the requested formulation. Namely,

$$\vec{n}(\vec{r}-\vec{r}_0)=0$$

is the normal equation of the plane, with $\vec{r}=<x, y, z>$ and $\vec{n}, \vec{r}_0$ appropriately defined above.

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Is this procedure correct? Thanks in advance.

Answer 1

The normal vector of the plane defined by three points is defined by

$$ \vec{n} = \vec{P}_0 \times \vec{P}_1 + \vec{P}_1 \times \vec{P}_2 + \vec{P}_2 \times \vec{P}_0 $$

The equation of the plane is therefore

$$ \vec{n} \cdot \pmatrix{x \\ y \\ z} = d $$

where $d = \vec{n} \cdot \vec{P}_0$

If you make $\vec{n}$ a unit vector, then $d$ is the minimum distance of the plane to the origin.

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Unrelated but interesting, the area of the triangle defined by three points is

$$ A = \tfrac{1}{2} \| \vec{P}_0 \times \vec{P}_1 + \vec{P}_1 \times \vec{P}_2 + \vec{P}_2 \times \vec{P}_0 \| $$