Implicit equation of a line in 3 Dimensions (disambiguation)?

Question

The implicit equation of a line in 2D is:

$Ax + By +C =0$

It's analogous in 3D is an implicit plane, described by:

$Ax+By+Cz+D=0$

What is then the implicit equation of a line described parametrically by:

$P=P_0+t\vec v$

In 3 dimensions?

Answer 1

You cannot write the line in 3D as the locus of a single linear equation.

Roughly speaking, each equation removes one degree of freedom from a space. So a system of equations containing only a single equation describes a hypersurface, an $(n-1)$-dimensional subset of $n$-dimensional space. So in 2D that gets you a curve. A line, if it's a linear equation. In 3D, it gets you a surface. A plane if it's linear. Etc in higher dimensions.

If you want to cut out a 1D curve from 3D space, you need a system of two equations, in the general case. And for a 1D line, it should be a system of linear equations, and of course, they must not be inconsistent or underspecified. Start with three degrees of freedom, remove two, and you have one remaining. A space with one degree of freedom is a curve. A line, if the equations are linear.

So there's no way to cut a line out of 3D space with a single linear equation. But parametric equations don't impose relationships on coordinates. They don't remove degrees of freedom. Instead, they take some degrees of freedom and map them into larger spaces. The parameters are still varying independently. So with parametric equations, you don't subtract a degree of freedom for each equation. Instead, you add a degree of freedom for each parameter. And for a parametrized equation like $P=P_0+tv,$ there is only one parameter. So it's a line, and this works in an affine space of any dimension. In particular, you can also represent the equation of a line in the plane this way.

So in summary, if you want a uniform equation of a line in all dimensions, it's $P=P_0+tv.$ If you want a uniform equation of an affine hyperplane in all dimensions, it's $A_1x_1+\dotsb+A_nx_n=D$. They're not the same thing, except in two dimensions where $n-1$ and $1$ agree. That's the coincidence that's throwing you off.

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As a postscript, in response to the comments, let me note that the claim, which goes by the name of "constraint counting", that a system of $k$ equations cuts out a $k$-codimensional (that is, $(n-k)$-dimensional) subspace of $n$-dimensional space is just a rule of thumb.

Morally it should be true, but there are some factors that can cause it to fail. These include if the equations are not independent, if the equations are inconsistent, if the number system is not algebraically closed, and non-transversal intersections.

So for example, making use of these pathologies, you can actually write the equation of a line parametrized as $(x,y,z) = (x_0,y_0,z_0) + t(a,b,c)$ in three dimensions as $$\left(y-y_0-\frac{b}{a}(x-x_0)\right)^2+\left(z-z_0-\frac{c}{a}(x-x_0)\right)^2=0.$$

While this does give the implicit equation of a line, and may therefore be taken as the answer to your question, morally I still think the right thing to say is that by dimension counting you should use 2 equations to describe a space of codimension 2.

Much mathematical machinery goes into making this fact which "wants" to be true, true. For example if you count the rank of a coefficient matrix of a system of linear equations, or the codimension of the Gröbner basis of a system of equations more generally, you can eliminate the dependent equations. If you projectivize your space, parallel lines intersect at infinity and some inconsistent equations gain solutions. You can just pass to the algebraic closure of your field of numbers, or else look at the $\bar{k}$-points of your variety over non-algebraically closed field $k$. And a structure sheaf can allow you to count intersections with multiplicity, or other derived point of view techniques, which make your non-transversal intersections better behaved.

Even without all that machinery, it will be the case generically that a system of $k$ equations cuts out a space of codimension $k$. Meaning that if you choose a system at random, you will obtain one with this property with probability one. And with the right abstractions, you can make it also true for most of the special cases.

And even with all that abstraction in place, there exist varieties which are not complete intersections, like the twisted cubic, a curve in three dimensional space which cannot be cut out by only two equations. But our rule of thumb is at least locally true even for them, if not globally.

Answer 2

A plane can be defined implicitly as:

$$\vec{x} \cdot \vec{n} = d$$

And it can be normalised by imposing the condition $\lVert \vec{n} \rVert = 1$.

While there is no scalar linear equation that describes a 3D line implicitly, there is a vector equation which is analogous.

Taking the parametric equation:

$$\vec{x} = \vec{p} + t\vec{d}$$

Which, again, we can normalise with $\lVert \vec{d} \rVert = 1$. We can eliminate $t$ by taking the cross product with $\vec{d}$:

$$\vec{x} \times \vec{d} = \vec{p} \times \vec{d} + t\vec{d} \times \vec{d} = \vec{p} \times \vec{d}$$

We define the moment vector to be:

$$\vec{m} = \vec{p} \times \vec{d}$$

Then an implicit equation of the line is:

$$\vec{x} \times \vec{d} = \vec{m}$$

The moment vector has a simple intuition behind it. Any line that does not pass through the origin defines a unique plane that includes both the line and the origin. The moment vector $\vec{m}$ is normal to this plane, and the length of $\vec{m}$ is the closest distance between the line and the origin. (Note that if $\vec{d}$ is not normalised, the minimum distance to the origin is $\lVert \vec{m} \rVert / \lVert \vec{d} \rVert$.)

To convert from the implicit equation to the parametric one, you need some point on the line. The most convenient one is the point closest to the origin, which is very simple to calculate:

$$\vec{p}_\bot = \vec{d} \times \vec{m}$$

This answer doubles as a stealth introduction to Plücker coordinates, which is worth knowing about.