Equation of plane -- point/vector pedagogy

Question

Suppose we have a point $\mathbf P$ and a vector $\mathbf n$ in plain ordinary 3D space. Here I am deliberately using upper-case letters for points, and lower-case points for vectors, since they are two different things. Then the set of points $\mathbf X$ that satisfy $(\mathbf X - \mathbf P) \cdot \mathbf n = 0$ is a plane, as we all know. The quantity $\mathbf X - \mathbf P$ is the difference of two points, so it's a vector, and taking its dot product with $\mathbf n$ is legitimate. So far, so good.

But, it's very tempting to rewrite this as $\mathbf X \cdot \mathbf n = \mathbf P \cdot \mathbf n$. Then, if I let $\mathbf P \cdot \mathbf n = d$, the plane equation becomes $\mathbf X \cdot \mathbf n = d$. From a programming point of view, this is nice -- I can now represent the plane by four numbers ($\mathbf n$ and $d$) rather than six ($\mathbf n$ and $\mathbf P$). Also, if I write $\mathbf P = (a,b,c)$, then the equation becomes $ax + by + cz = d$, which is the plane representation that we all know from high school geometry. Nice, comfy, familiar. Good.

The problem is that expressions like $\mathbf X \cdot \mathbf n$ and $\mathbf P \cdot \mathbf n$ don't make sense -- everyone knows that you can't take dot products of points and vectors. Computationally, everything works fine, but pedagogically, it feels like something is wrong.

Can anyone make some sense out of this, please? I'd like to write plane equations in the form $\mathbf X \cdot \mathbf n = d$ and still be able to sleep peacefully at night..

Answer 1

It's fine. A point $X$ is usually identified with the vector $X-O$, where $O=(0,0,0)$ is the origin, and therefore one usually writes $X$ for both the vector and the point. If you still insist on not using the same notation for both, you can simply subtract $O$ on both sides:

$$(X-P)\cdot n = 0 \iff ((X-O)-(P-O))\cdot n=0 \iff (X-O)\cdot n=(P-O)\cdot n$$

Answer 2

I guess this depends on the way things are defined. Some guys say that the space is $\mathbb{R}^3$ from the start, and for them points and vectors are the same. They sleep at night very peacefully I guess.

Then again, others appreciate a distinction between affine spaces and vector spaces. This distinction is there throughout school geometry. In this case, you'll probably have to randomly select an origin $O$ and instead of $X \cdot n$ write $(X-O) \cdot n$.

But, there's a third option! You can sort of make sense of the notation $X \cdot n$ and still respect the fact that affine spaces don't have a starting point. Problem is, you probably shouldn't show this to children, otherwise they might become the ones losing their sleep. Also, you won't be able to compare values $X_1 \cdot n_1$ and $X_2 \cdot n_2$ if $n_1 \neq n_2$.

You have probably already guessed where I'm driving at. Suppose $n$ is a vector. Define an equivalence relation $\sim_n$ like this: $X \sim_n Y$ iff $(Y-X)\cdot n = 0$. Then, if $A$ is the whole affine space and $V$ is the set of its vectors, then $\sim_n$ is an equivalence relation on $A$ and $A/\sim_n$ has a natural structure of an affine space (whose space of vectors is $\langle n \rangle$). Then you can simply denote by $X \cdot n$ the class of $X$ with respect to equivalence $\sim_n$.

With this notation, $X \cdot n = const$ really defines a plane, with the only drawback that the constant on the right is not a number, but rather just an element of a one-dimensional something without any point of reference. Now that I think about it, this constant is the plane itself ))

This is mostly useless I guess, except maybe for the sleeping problem.