My question is very elementary; I just want to ask if it is widely known (probably yes) and whether this is written in textbooks (where). A line in the 3-dimensional space is usually given either by two points or by a point and a vector; in total 6 numbers. However, 5 numbers are enough. As the direction vector (3 numbers) is necessiraly non-zero, we can take its first nonzero component and specify only 2 point coordinates (while the third, corresponding to the non-zero vector component under consideration, is fixed - say, 0).
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For more information read the Wikipedia article Plucker coordinates. The total number is $6$, $5$, or $4$ depending on how you count. – Somos Apr 14 '19 at 11:45
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@Somos: I believe that 5 numbers are enough in every situation. The question is whether 4 numbers are enough. It follows from the discussion below that "almost" yes: but it seems that some extra information is necessary. – Quiriacus Apr 15 '19 at 08:53
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The key point here is that the space of lines in 3D is four dimensional. Introducing coordinates for this space is not trivial. Consider the situation for a sphere in 3D. It is a 2D surface so two coordinates should be enough but how? Longitude and latitude? The north and south pole are exceptional points. In general you need an atlas and its charts. – Somos Apr 15 '19 at 10:30
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@Somos: I understand. However, I think it corresponds to my response to the answer of Alex Ravsky (it wants some additional information, which of the coordinates is considered as "critical"). A way of a construction of an atlas is not given implicitly (or yes?). It needs some additional information - "to know where to start"... – Quiriacus Apr 16 '19 at 10:59
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Better to specify your direction vector with $\theta$ and $\varphi$ in spherical coordinates $(r,\theta,\varphi)$. Otherwise you have a problem with how to specify which of the three coordinates is omitted $-$ this requires another parameter.
But in fact a line only has four degrees of freedom, not five. This is because you can choose any point on the line to represent it. Actually finding four numbers to represent a line is a bit messy; start with the direction $(\theta,\varphi)$, and consider the plane $L$ through the origin that is perpendicular to it. Then the line is specified by its point of intersection with $L$. This requires only two coordinates, but you need some linear algebra to set up your coordinate system on $L$.
Perhaps somebody knows a more elegant parametrisation?
TonyK
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Equivalently, fix a pair of orthogonal planes. The line can be parameterized by its intersections (perhaps at a point at infinity) with those planes—4 degrees of freedom. – amd Apr 14 '19 at 20:07
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@amd: OK, but that is another messy solution: how do you specify the point at infinity? – TonyK Apr 14 '19 at 20:14
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I suspect that there’s a tradeoff between convenience and parsimony in this—anything that tries to stick to the minimal 4 parameters will end up being “messy.” – amd Apr 14 '19 at 20:46
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@amd: Both planes cannot go through the origin. We need some more information. – Quiriacus Apr 15 '19 at 04:57
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@Quiriacus The problem isn’t the origin, but the planes’ line of intersection. For any pair of fixed planes, lines that pass through that intersection can’t be uniquely specified by their intersections with the planes. – amd Apr 16 '19 at 17:02
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Even though there are four degrees of freedom, I doubt that there’s a “nice” parameterization that uses only four numbers. I suspect that there is always some choice to be made in order to interpret them: there’s an algorithm to identify a line given the four values, but there’s not a “nice” single formula. For example, in your proposal, a basis or reference direction must be chosen for the perpendicular plane in order to interpret two of the parameters, and that choice depends on the other two. – amd Apr 16 '19 at 17:11
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The idea is right, but, in general, we cannot fix a given coordinate of the direction vector to be $0$ or $1$. So usually we assume that the direction vector $(d_1,d_2,d_3)$ has unit length, that means that $d_1^2+d^2_2+d^2_3=1$.
Alex Ravsky
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1This is a misunderstanding. I don't fix any direction vector coordinate. But I say that I need 3 numbers from the direction vector and 2 numbers from the point. (The point is lying on a fixed plane, e.g. on w=0, where w is x or y or z.) – Quiriacus Apr 14 '19 at 11:08
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@Quiriacus OK, then we indeed always can assume that there exists a coordinate of the direction vector which is equal to $1$. – Alex Ravsky Apr 14 '19 at 11:11
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@Quiriacus I don't understand why $4$. We have three numbers for a point and three other numbers for a vector, among which one equals $1$. – Alex Ravsky Apr 14 '19 at 11:36
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14 numbers $a,b,c,d$ are enough to determine a line from the direction vector $(1,a,b)$ and the point $[0,c,d]$. But it wants some additional information, which of the coordinates is considered as "critical" (it is $x$ in my example). – Quiriacus Apr 15 '19 at 05:09
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The specification of lines in three dimensional Euclidean space by four numbers is not trivial. Let us start with a Euclidean plane. We can construct a Cartesian coordinate system where each point is uniquely specified by two coordinates. What about lines? Without projective coordinates, there is a difficulty. For example, consider two reference lines $L_0$ and $L_1$ with $L_0$ given by the first coordinate being $0$ and $L_1$ by the first coordinate being $1$. Then, any line $L$ not parallel to $L_0,L_1$ will intersect the two reference lines and the second coordinates of the points of intersection will determine $L$ uniquely. The only exceptions are lines parallel to $L_0,L_1$. In that case, the first coordinate determines the line uniquely. That is, $L_x$ is the line all of whose points have first coordiante $x$.
We can extend this to three dimensional Euclidean space. We construct a Cartesian coordinate system where each point is uniquely specified by three coordinates. Consider two reference planes $P_0$ and $P_1$ with $P_0$ given by the first coordinate being $0$ and $P_1$ by the first coordinate being $1$. Then any line $L$ not parallel to $P_0,P_1$ will intersect each of the two reference planes and ech of the two points of intersection are given by coordinates of which the first coordinate is already fixed, and the remaining two coordinates of each point uniquely specify the line $L$ with four numbers. The only exceptions are the lines parallel to $P_0,P_1$. In that case, the first coordinate specifies a plane that contains the line $L$ and in that plane, by the case for lines in a plane, that is uniquely determined by two numbers, with exceptions noted already.
In summary, there seems to be no simple way to always uniquely specify every line in three dimensional Euclidean space with four real numbers without exceptional cases where less than four numbers are used. The fundamental reason for this situation is that the space of lines is topologically not the same as four dimensional Euclidean space. This is similar to why a circle is not topologically the same as a line, but if you remove one point on the circle, then it is homeomorphic to a line.
Somos
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