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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..

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The space of points of $\mathbb{R}^n$ and space of vectors of $\mathbb{R}^n$ is the same space (the isomorphism is $v \mapsto v+0$). Treat everything as vectors and you should be fine ;-) –  dtldarek Mar 11 '13 at 13:10
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I thought the space of points was an affine space, and the space of vectors was a vector space. These are different animals, aren't they? –  bubba Mar 11 '13 at 13:13
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Those are different only in the way that there is no "zero point" in affine space as there is the zero vector in vector space. However, after the introduction of coordinates, the origin becomes "the zero point" and in fact you are using vectors, not points. Still, with no origin the affine space and vector space (which does have origin by definition) are, as you pointed out, different species. –  dtldarek Mar 11 '13 at 13:31
    
The comment about coordinates makes sense. As soon as I introduce coordinates, there is automatically a point with coordinates $(0,0,0)$. So, now, $(a,b,c)$ is just shorthand for $(a,b,c) - (0,0,0)$, which is a vector. Good. But, $\mathbf X \cdot \mathbf n$ is coordinate-free, isn't it? –  bubba Mar 11 '13 at 13:43
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The dot product implies the origin, i.e. the unique point for which $a \cdot a = \|a\|^2 = 0$. –  dtldarek Mar 11 '13 at 14:05
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2 Answers 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.

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Certainly not useless. Thanks. –  bubba Mar 11 '13 at 13:54
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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:

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

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I'm not comfortable with the "$+$" in $X-O-P+O$, but I'd be OK with $(X-O) - (P-O)$. –  bubba Mar 11 '13 at 13:19
    
Fair enough. Edited. Is this okay for you? Otherwise I will need you to give me your definitions of points and vectors in $\mathbb R^3$. In linear algebra, one usually uses vectors to refer to points, and the distinction is only a semantical one, so there is no controversy in taking dot products between points and vectors. –  Samuel Mar 11 '13 at 13:23
    
I have a feeling that lack of decent definitions might be the root cause of the problem. If points and vectors only differ semantically, how do I add two points together? By adding the corresponding vectors? But won't the answer depend on what origin I choose? –  bubba Mar 11 '13 at 13:39
    
You need to provide definitions of points and vectors, otherwise your question has no answer. You say you're only talking about $\mathbb R^3$, so you don't need to talk about affine spaces or abstract vector spaces or charts of manifolds or equivalence classes. $\mathbb R^3$ has a canonical coordinate system, with $(0,0,0)$ as the origin. Then one can either say that $\mathbb R^3$ consists of points, and define vectors as arrows from the origin to a point, or one can start with vectors and define points as the endpoint of a vector that starts at the origin. –  Samuel Mar 11 '13 at 23:21
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@bubba: ...and once one has done that, there is an obvious isomorphism between your vector space and your space of points, so when you write $X\cdot n$ you actually mean $(X-O)\cdot n$. The dot product is only defined for vectors, and so is addition. Note that in writing $(X-O)\cdot n=(P-O)\cdot n$ you don't need to add points, and the equation remains true if you change your origin (although the values will change). –  Samuel Mar 11 '13 at 23:24
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