0
$\begingroup$

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.


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.


Is this procedure correct? Thanks in advance.

$\endgroup$
3
  • $\begingroup$ Being normal to the plane is not the same as being normal to $\vec{v}$. $\endgroup$
    – angryavian
    Commented Nov 18, 2022 at 1:54
  • 1
    $\begingroup$ I would take the cross product of $P_1-P_0$ and $P_2-P_0$ $\endgroup$ Commented Nov 18, 2022 at 3:22
  • $\begingroup$ That makes sense. Leaving that aside, should the normal vector have been defined as the cross product of those two vectors, the rest of the formulation would have been correct? Or are there any other errors? $\endgroup$
    – lafinur
    Commented Nov 18, 2022 at 12:31

1 Answer 1

1
$\begingroup$

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.


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

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .