3-Dimensional planes

Notes:

Equation of a Plane
Couple sections ago, we saw that the equation of plane can be expressed as $ax+by+cz=d$. However, this equation does not give much information. So suppose we have the following graph:

Where $\vec{r}$ and $\vec{r_0}$ are position vectors for points $P$ and $P_0$ respectively, and $\vec{n}$ is a normal vector that is orthogonal (perpendicular) to the plane.
Since $\vec{r} - \vec{r_0}$ is on the plane, then $\vec{n}$ is orthogonal to $\vec{r} - \vec{r_0}$. In other words, their dot products should give 0.
So,

$(\vec{r} - \vec{r_0}) \cdot \vec{n} = 0 \to (\lt x,y,z\gt - \lt x_0,y_0, z_0\gt) \cdot \lt a,b,c\gt = 0$
$\to \lt x-x_0 , y-y_0, z-z_0\gt \cdot \lt a,b,c\gt = 0$
$\to a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$

Which is formula for the equation of a plane.

The key to finding the equation of a plane is finding two things:

1. The normal vector (orthogonal to the plane)
2. A point on the plane.

Then you can just plug those into the formula to get the equation!