# How many degrees of freedom does an $n$-dimensional line have?

I know that for a two dimensional line, there are two degrees of freedom $$(\theta, d)$$ where $$\theta$$ is the angle that the normal makes with the $$x$$ axis, and $$d$$ is the distance from the line to the origin.

For a three dimensional line, I was reading this question here: 4-dof of a 3d line

The argument for $$4$$ degrees of freedom for a line in $$3$$d that I understood the best says that each line is tangent to a unique sphere of radius $$r$$, intersecting at a point $$m = (r, \theta, \phi)$$ in spherical coordinates. Then the angle that the line's direction vector makes with the ray from the center of the sphere to the point $$m$$ is the final degree of freedom.

I'm wondering if there is a generalization to lines of $$n$$ dimensions. I would be tempted to say it's $$n+1$$ degrees of freedom, however this isn't true for $$n=2$$. Any insights appreciated.

• You have $n-1$ degrees of freedom for the unit direction vector $\vec v$ of the line and another $n-1$ degrees of freedom for the point in the hyperplane orthogonal to $\vec v$ through which the line passes. So the answer is $2(n-1)$. [This reasoning is precisely analogous to your reasoning in the case $n=2$.] Commented Apr 4, 2023 at 21:39

To define a line in $$\mathbb{R}^n$$, we need two points. This gives $$2n$$ degrees of freedom.
But for one line each point can be any point in the line. This substracts one degree of freedom for each point, so we have $$2n-2$$ degrees of freedom.
Another way: in $$\mathbb{R}^n$$ the line direction has as many degrees of freedom as (half of) the unit sphere $$S^n$$, i.e. $$n-1$$ because it is an hyper-surface.
Then the set of lines that have a given direction is in bijection with an hyperplane orthogonal to that direction. This adds $$n-1$$ degrees of freedom, so total is $$2n-2$$.
• @IntegrateThis For each of the $2$ points, you can slide it as you want on the line without changing the line. So we subtract the degree of freedom the point has on the line, which is $1$. There are $2$ points to choose so we subtract $2$. Commented Apr 4, 2023 at 22:09
• @IntegrateThis Or, if you prefer: the set of pairs of points in $\mathbb{R}^n$ has dimension $2n$. The set of pairs of points on a line has dimension $2$. So the set of lines in $\mathbb{R}^n$ has dimension $2n-2$. Commented Apr 4, 2023 at 22:14