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The equation of a line in $ \ \mathbb{R}^2 \ $ can concisely be expressed as $ \ ay+bx=c \ $. (And similarly, a plane in $ \ \mathbb{R}^3 \ $ is expressed as $ \ ay+bx+cz=d \ $).

I've tried to find a similar way to express a line in $ \ \mathbb{R}^3 \ $ but I can't find any. The closest I got was the vector equation of a line, but that's too dissimilar from $ \ ay+bx=c \ $, and also a bit more complicated.

Am I correct in thinking there isn't any similar way to express a general 3-D line in this way? (Sorry for the vagueness of the word "similar").

Stephen
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4 Answers4

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If $f(x, y, z)$ is a differentiable function (with the derivative satisfying some special conditions*) of 3 variables x, y, z (it can end up not depending on some of those variables), then the equation $f(x, y, z) = c$ defines a 2-dimensional "manifold" (think of it like a curvy sheet of paper).

In general, if f is a function of n variables, then f(those n variables) = c defines an n-1 dimensional manifold.

A plane is a 2 dimensional manifold, which is why you can express it the way you did. A line is a one dimensional manifold, which is why we can't just express it with something like $ax+by+cz=d$—we need more equations to reduce the dimension of the manifold it defines. (hence the vector formulation!)

A good way to think about manifolds is that the number of variables are the degrees of freedom, and the number of equations are the constraints, and the dimension of the shape you're producing is degrees of freedom minus constraints.

For a plane in 3-dimensional space, you have 3 variables and one equation. 3-1=2, which is the dimension of a plane. For a line in 3-space, we have 3 variables, but a line is one-dimensional, so one equation does not suffice to describe it.

*If you're interested, this "special condition" is surjectivity. If we have this property at a point, then the point set is locally a manifold.

**If you're really interested, look into how one can use something called the implicit function theorem (which is, among other things, a way of stating the relationship between degrees of freedom and constraints) to define manifolds. Vector Calculus, Linear Algebra, and Differential Forms by Hubbard & Hubbard gives an excellent and accessible treatment of the subject.

BRSTCohomology
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  • You're right. I didn't want to go into the conditions for being a differentiable manifold, but I see my simplification was actually misinformation. I'll fix it. – BRSTCohomology Jan 02 '18 at 04:09
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The most natural way to describe a line in $\mathbb{R}^3$ as a solution set is using two linear equations. Namely, take two planes whose intersection is your line, and take their equations. So any line in $\mathbb{R}^3$ can be described by a system of equations of the form $$ax+by+cz=d$$ $$ex+fy+gz=h.$$

It actually is possible to combine these two equations into a single equation, by a somewhat artificial trick. Namely, you use the fact that if $s,t\in\mathbb{R}$ then $s=t=0$ iff $s^2+t^2=0$. In this case, we take $s$ and $t$ to correspond to our pair of equations, and thus get a single equation $$(ax+by+cz-d)^2+(ex+fy+gz-h)^2=0$$ whose solution set is our line. (This single equation is usually no easier to use than the system of two equations though; it is mainly of interest just to illustrate that it is possible to use a single polynomial equation.)

Eric Wofsey
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Although there are ways to represent a line in 3-d with a single equation, for instance, as a degenerate quadric, there’s no way to represent it as a single linear equation. There is, however, a way to convert a parametric equation for a line into a system of linear equations that can be written in a succinct form.

Let the line be parameterized as $\mathbf p + \lambda \mathbf v$. Expand by coordinates and solve for $\lambda$: $${x - p_x \over v_x} = {y-p_y \over v_y} = {z-p_z \over v_z}.$$ This only works if none of the components of $\mathbf v$ is zero. If it is, then you will need a separate equation that sets the corresponding variable to a constant value.

amd
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You want a line, which will of course take the form: $$\left\{\begin{bmatrix}a+b\lambda\\c+d\lambda\\e+f\lambda\end{bmatrix}\mid \lambda\in\Bbb R,\quad a,b,c,d,e,f\in\Bbb R\right\}$$$$=\left\{\begin{bmatrix}a\\c\\e\end{bmatrix}+\lambda\begin{bmatrix}b\\d\\f\end{bmatrix}\mid \lambda\in\Bbb R,\quad a,b,c,d,e,f\in\Bbb R\right\},$$ i.e. $x(\lambda)=a+b\lambda, y(\lambda)=c+d\lambda,z(\lambda)=e+f\lambda$

Or as symmetric equations, if $b,d,f$ are all nonzero: $$\frac{x-a}{b}=\frac{y-c}{d}=\frac{z-e}{f}=\lambda$$