4-dof of a 3d line Degrees of freedom of a line if $R^3$ sort of confuse me.  I read that it has 4 dof.  The text proposes a way to count these dof: think of two perpendicular planes s.t. the intersection of a line with each of the planes constrains 2 parameters.
The question is, what about the lines that cross only one of these planes, how am I supposed to think about them?
Another question is: a line is defined by a pair of points.  Each point has 3dof taken by itself and it seems that when one of them is chosen, there's no constraint on choosing the 2nd one, so why aren't there 6dof?
[Edit] Could you please also provide a definition of what DOF?  Is it  a property of a point set?
 A: One way to count the degrees of freedom is to note that a line is defined by two points (six degrees of freedom), and is invariant under translation of either of the two points along the line so defined (subtracting two degrees of freedom, leaving four).
Another way is to define the line by a single point (three degrees of freedom) and a direction (two more degrees of freedom), with invariance under translation of the point along the line so defined (subtracting one degree of freedom, leaving four).
A third way is to note that any line is tangent to a unique sphere centered at the origin.  The line can be defined by the radius of the sphere (one d.o.f.), the point at which the line meets the sphere (two d.o.f.), and the direction in which the line leaves from the sphere (one more d.o.f., making four).
A: The given explanation neglects lines crossing at most of the two planes (and I assume the text uses parallel instead of perpendicular planes), that is those lines parallel to these two planes. You can do the same with two other pairs of parallel planes so that each line is parallel to at most one such pair, i.e. at least one of the three methods can be used for the line. The union fo finitely many sets with four degrees of freedom each is itself a set with four degrees of freedom.
A line can be given by two (distinct) points $A,B$ on the line, so that would suggest six degrees of freedom indeed. However, you get the same line if you translate $A$ and/or $B$ along the line, which takes away two degrees of freedom again.
Alternatively, for a line $\ell$ let $P$ be one of the two intersection points of the unit sphere with the line thruogh $O$ parallel to $\ell$ and let $Q$ be the pint where $\ell$ intersects the plane $\pi$ perpendicular to $OP$ through $O$.
Then $P,Q$ determine $\ell$ again and we have two degrees of freedom for the choice of $P$ on th eunit sphere, plus two degrees of freedom for the choice of $Q$ in the plane $\pi$.
