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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.

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    $\begingroup$ 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 $n=2$.] $\endgroup$ Commented Apr 4, 2023 at 21:39

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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 $2n-2$.

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  • $\begingroup$ I understand the second way better, if you can elaborate why do we subtract a single degree of freedom for point knowing we can choose any two points on the line? $\endgroup$ Commented Apr 4, 2023 at 21:52
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    $\begingroup$ @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 $2$. $\endgroup$ Commented Apr 4, 2023 at 22:09
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    $\begingroup$ @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 $2n-2$. $\endgroup$ Commented Apr 4, 2023 at 22:14

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