What is the minimum information needed to determine a line in 3D?

Question

Motivation: A Line in $\Bbb R^2$

Any line can be uniquely determined by two points. In $\Bbb R^2$, a point is uniquely determined by two values (its $x$ and $y$ coordinates). Hence, to uniquely determine a line in $\Bbb R^2$ requires at most four values (in this case $x_0$, $y_0$, $x_1$, and $y_1$).

Those aren’t the only four values that can be used; we could also express the line parametrically, as the pair of equations:

$$x=x_0+at\\y=y_0+bt$$

Now the four values are $x_0$, $y_0$, $a$, and $b$. We can rearrange that pair of equations to get the symmetric-form equation of the line, which also determines the line by the same four values:

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}$$

No doubt most of you see where I’m headed, though. We can rearrange this to get the good old slope–intercept form of the line’s equation:

$$y=mx+c$$

This uniquely determines the line with just two values ($m$ and $c$)!

Some lines (namely those parallel to the axes) can be determined with just one value, but two is the minimum information necessary to generally determine a line in $\Bbb R^2$—not four, after all.

The Problem: A Line in $\Bbb R^3$

A line in $\Bbb R^3$ is still able to be determined by two points, although each point now needs three values ($x$, $y$, and $z$). Hence, to uniquely determine a line in $\Bbb R^3$ requires at most six values.

These six values can be $x_0$, $y_0$, $z_0$, $x_1$, $y_1$, and $z_1$; or in parametric form…

$$x=x_0+at\\y=y_0+bt\\z=z_0+ct$$

…or symmetric form…

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$

…they could be $x_0$, $y_0$, $z_0$, $a$, $b$, and $c$. But no matter where I’ve looked or what I’ve tried, I can’t do it in less than six. Can it be done?

Worth noting is that a plane in $\Bbb R^3$ can be determined in fewer than the naive nine values (three points): using one point and a normal vector (three values each), we can do it in six. It seems counterintuitive that a plane and a line should both be determined by six values… but then, the same could be said about a line and a point in $\Bbb R^2$.

Answer 1

Ignoring lines passing through the origin for the moment, a given line has a unique point nearest to the origin (and this point can be determined using three parameters $x_0,y_0,z_0$). After this, it remains to pick a direction for the line within the plane that is orthogonal to the line through $(0,0,0)$ and $(x_0,y_0,z_0)$. While it is not straightforward to do this in a "uniform" way, such a direction requires only one parameter. So in total, four parameters suffice.

More generally, a line (in general position) in $\Bbb R^n$ is determined by a point in $\Bbb R^n$ and a point on $S^{n-2}$, so by $2n-2$ parameters.

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Inspired by Jyrki Lahtonen's comment: We may look at the matter this way: The naive approach is to use two points $(x_1,\ldots, x_n)$ and $(y_1,\ldots, y_n)$ on the line (thus $2n$ parameters). However, the positions of the points within the line are irrelevant (except that they must be distinct). As such we "waste" one parameter per point and should be able to get along with $2n-2$ parameters. To this end, we may pick specific points: There must exist at least one $k$ such that $x_k\ne y_k$. Then the points on the line assume all real values as $k$th coordinate. Hence it is possible to relocate $x$ and $y$ such that $x_k=0$ and $y_k=1$. Thus all it takes is $2n-2$ real-valued parameters together with a finite-valued parameter $k$.

The necessity of such a finite-valued parameter arises already in two dimensions: While we have $y=mx+b$ for "most" lines (with two parameters $m,b$), we need to cover also $x=c$ (with one parameter $c$) and need to specify which of the two forms we refer to in the first place.

Answer 2

It can be done with 5 values, 1. A point, that takes 3 values 2.one angle on the xy plane, 3.one angle on the yz plane

pretty much it is spherical coordinates with centre given at a point,

Answer 3

Here's a uniform way using $4$ parameters that is not only more intuitive but has a nice continuity property in the sense that two lines that differ by a small rigid perturbation will have only slightly different encodings.

Take any line $L$ in $3$-dimensional Euclidean space. We shall encode $L$ as a $4$-tuple $(p,q,r,s)$ where $p,q$ are angles and $r,s$ are positions. Let $(p,q)$ encode the rotation $R$ needed to orientate the $z$-axis to be parallel to $L$, where $R$ is the tilt in the direction of $p$ of angle $q$ when looking at the $xy$-plane. (More precisely, $p$ specifies a line in the $xy$-plane around which $R$ is a rotation of angle $q$. Note that this is not the same as polar coordinates.) Let $P$ be the image of the $xy$-plane under $R$. Let $(c,d)$ be the point where $L$ intersects $P$ from the perspective of $P$. (More precisely, let $(c,d)$ be the inverse image of $L \cap P$ under $R$.)

This encoding is continuous in the sense that a slight rigid perturbation (translation and rotation) only changes the encoding slightly. This is not normally possible if you want to encode a ray due to its direction, but for lines the above encoding is continuous. For rays, the only discontinuity is for rays that point in directly opposite direction to the $z$-axis, since slight perturbation of these rays would result in rays with all possible $p$.

Anyway, this construction generalizes to lines in higher dimensional spaces. You use the encoding for the direction in $n$ dimensions to specify the axis of the tilt in $(n+1)$ dimensions.

Incidentally, this encoding of direction that I have described is exactly what GLMol uses when one mouse-drags to rotate the diagram. The benefit is that a single mouse-drag will not be able to cause rotation around the axis pointing out of the screen, which is a major annoyance in many other 3d viewers such as JSMol (see this website where you can try both GLMol and JSMol).