Planes

In the context of linear algebra it’s important to have some knowledge about planes.

The general formula for a plane in 3D space is

Derivation

We will derive this formula in a few simple steps using some of the knowledge acquired so far. We start with some assumptions:

We suppose that we have…

• …a known point $P_0 = (x_0,y_0,z_0)$ which lies in the plane.

• …a point $P = (x,y,z)$ which represents any possible point in the plane.

• …a known vector that is orthogonal to the plane denoted by $\vec{n} = (a,b,c)$. We refer to this vector as the normal vector.

We then define the vectors $\vec{r}_0$ and $\vec{r}$ to be the position vectors pointing to the points $P_0$ and $P$ respectively.

A position vector of a point $P$ is the vector from the origin to $P$.

Also, we know that the vector $\vec{r} - \vec{r}_0$ lies inside of the plane, since $P$ and $P_0$ - which are the points pointed to by the vectors $\vec{r}$ and $\vec{r}_0$ respectively - both lie in the plane as well. The vector $\vec{r} - \vec{r}_0$ is the vector which starts at $P_0$ and points to $P$.

Since $\vec{n}$ is orthogonal to the plane, and $\vec{r} - \vec{r}_0$ lies in the plane, $\vec{n}$ must also be orthogonal to $\vec{r} - \vec{r}_0$.

If you recall from the lesson about orthogonal vectors, we know that the dot product of two orthogonal vectors is $0$.

Therefore

We can rearrange this as

We now write the dot products in coordinate form:

Note that $ax_0$, $by_0$, and $cz_0$ are all known values, so usually it is said that

and then the above equation can be rewritten as

And in fact, this is just the coordinate form of the equation, and our starting point

is the vector form of the same idea.

The vector form has a straightforward geometric interpretation. It’s basically stating that we are interested in all points $P$ such that the vector $\vec{r} - \vec{r}_0$ - the vector from $P_0$ to $P$ - is orthogonal to the normal vector $\vec{n}$.

Now if you notice, $a$, $b$, and $c$ are the components of $\vec{n}$. In fact, if you’re asked to find the normal (orthogonal) vector to the plane, and the plane is given to you in the form $ax + by + cz = d$, all you have to do is collect the components of $\vec{n}$ into a vector, and you’re done.

For example, take the plane

Just from looking at this, I know that the vector orthogonal to the plane is

Finding the Equation of the Plane From Three Points

Before we can get into the explanation itself, we must first go on a small digression.

Vectors From Points

When we want to express the idea of a vector which points from some point $A$ to some other point $B$, we can use the notation $\overrightarrow{AB}$.

The vector $\overrightarrow{AB}$ can be computed as

where $\vec{r}_B$ and $\vec{r}_A$ are the location vectors of the points $B$ and $A$ respectively.

The location vectors are simply vectors which point to some point. For example, the location vector of the point $C = (1,2,3)$ is

Back to Our Main Endeavor

It’s easy to find the equation of a plane when given one point and the normal vector, since we know that

and the left-hand side only needs $a$, $b$, and $c$. These are all values which come from the information we have directly.

But if we’re only given three points, there are some extra steps involved.

We can use one point for our $x_0$, $y_0$, and $z_0$ values. We then combine the three points to find two vectors in the plane.

If we’re given three points

We now want to find two vectors lying in the plane. To do this, we compute the vectors $\vec{P_2 P_1}$ and $\vec{P_2 P_3}$ as described above:

We now know that our normal vector must be perpendicular to both of these vectors, thus

and

We have a linear system of equations:

To solve this in terms of $c$, we can use any method we want. I will employ substitution:

We can choose any value for $c$, say $c = 1$, then

So now,

For $x_0$, $y_0$, and $z_0$, we can pick any one of the given points, say $P_1$.

Our equation now becomes:

which is the equation for our plane.

Checking the Solution

We can even by using the points given as our $x$, $y$, and $z$:

For $P_1$:

For $P_2$:

For $P_3$:

A Final Remark on the Normal Vector

It’s important to note that the normal vector cannot be all zeroes, so

This is because - as outlined in the lesson about orthogonal vectors - by definition, the zero vector $\vec{0}$ is orthogonal to all vectors, and it not uniquely determine a plane.