
We now address the question: what is the relation between the different ways to describe a plane  by points, one point and a vector, or by an equation?
Suppose the points ${P}_{1},{P}_{2},{P}_{3}$ lie in plane $Q$ and they are not all on a line.
Then the vectors $\stackrel{\u27f6}{{P}_{1}{P}_{2}}$ , and $\stackrel{\u27f6}{{P}_{1}{P}_{3}}$ have direction in $Q$ and an arbitrary point in $Q$ will have the coordinates of $\stackrel{\u27f6}{O{P}_{1}}+s*\stackrel{\u27f6}{{P}_{1}{P}_{2}}+t*\stackrel{\u27f6}{{P}_{1}{P}_{3}}$ for some pair of values $(s,t)$ .
This is called a "parametric" representation of the plane with parameters $s$ and $t$ .
( $s$ and $t$ can be considered the components of the point in the plane in the basis given by $\stackrel{\u27f6}{{P}_{1}{P}_{2}}$ and $\stackrel{\u27f6}{{P}_{1}{P}_{3}}$ with origin ${P}_{1}$ .)
You can compute a normal to $Q$ by taking the cross product $\stackrel{\u27f6}{{P}_{1}{P}_{2}}\times \stackrel{\u27f6}{{P}_{1}{P}_{3}}$ .
We abbreviate by defining
and so the equation of the plane becomes
where $\stackrel{\u27f6}{OP}=(x,y,z)$ .
You can write this out explicitly as
It is common, but not necessary to "normalize" $\stackrel{\u27f6}{N}$ , that is to replace it by $\stackrel{\u27f6}{n}$ with $\stackrel{\u27f6}{n}=\frac{\stackrel{\u27f6}{N}}{\left\stackrel{\u27f6}{N}\right}$ (remember $\left\stackrel{\u27f6}{N}\right={\left(\stackrel{\u27f6}{N}\xb7\stackrel{\u27f6}{N}\right)}^{1/2}$ ) here.
In practice, planes are usually described by a normal vector, like $\stackrel{\u27f6}{n}$ here, and by a point in it.
We started with three points, and obtained a parametric representation of the plane from them. We then found an equation describing the plane from that representation.
If we can go from the description of $Q$ by this equation back to three points in $Q$ , we will be able to go all the way around the circle and find any representation of $Q$ from any other.
There are an infinite number of points in $Q$ and choosing three of them requires making arbitrary decisions to single out three of them.
If $\stackrel{\u27f6}{N}$ has all three of its components nonzero we can set each pair of variables to zero and solve for the third one. Then the three points will be
and
which are the points at which the plane meets the three axes.
From these points you can go around the circle again, and determine any representation of $Q$ .
In the applet here you can enter three arbitrary points, and it will find and picture the plane, show $\stackrel{\u27f6}{N},(\stackrel{\u27f6}{N},\stackrel{\u27f6}{O{P}_{1}})$ , and the parametric representation of points on it. You can do all of these things except making the picture, yourself.
Exercises:
5.2 Write down the equation for the default plane in this applet, and find the three points in that plane which have two 0 coordinates each.
5.3 Start with three random points and go through this procedure to find $\stackrel{\u27f6}{N}$ and the points where the plane meets the axes.
5.4 Set up a spreadsheet that does this whenever $\stackrel{\u27f6}{N}$ has all its components nonzero.
When ${N}_{z}$ is not zero, we can solve the equation of the plane, $\stackrel{\u27f6}{N}\xb7\stackrel{\u27f6}{OP}=\stackrel{\u27f6}{N}\xb7\stackrel{\u27f6}{O{P}_{1}}$ for $z$ in terms of $x$ and $y$ , getting
The coefficients of $x$ and $y$ here are particularly interesting to us. If you fix $y$ , then our three dimensional space becomes a plane. $\frac{{N}_{x}}{{N}_{z}}$ then represents the slope of the line that is the intersection of our plane and the plane described by the equation: $y=\text{constant}$ . The same statement holds after interchanging $x$ and $y$ in this one.
5.5 Find the two slopes (with $y$ fixed and $x$ fixed) h for the planes you describe in exercises 5.3 and 5.4.
