# Understanding distance from a line to a point at infinity (Jaan Kalda's handouts)

I'm reading page-7 of Jaan Kalda's Kinematics handouts, the following is written:

Here we can use a small addition to the last idea: if the quickest way to a plane (in a 3-d problem) or to a line (in 2-d) is asked then this plane or line can be substituted with a point very far (at infinity) in the perpendicular direction to it. The reason for that is quite simple: it takes the same amount of time to reach any point on the plane (line) from that very-very distant point. If we think about this in terms of geometrical optics then it means that a set of light rays normal to the surface falls onto the plane (line).

I'm having a bit of trouble how in the limit of taking the point far away that the distance from a line to this point is same. Could someone explain the above idea using physical examples?/Elaborate Kalda's example of light rays at the end?

I'm not entirely sure what the author's intent was here, but it seems like it's just a somewhat convoluted way to say that the shortest distance to a line/plane is along a line that's perpendicular to said line/plane - which then constrains one of the angles in the problem that's being discussed.

As for a physical example, consider sunlight. Since the Sun is so far away, rays of sunlight are often described as essentially parallel. For the same reason, the distance to the Sun from any two points on Earth can be taken to be the more or less the same. If you replace the Sun with a distant star, the error becomes minuscule. Now, a distance to a "point at infinity" is not well-defined, but in the limit, the error vanishes. Aiming at a point at infinity just gives you a direction (and it's the same direction everywhere, from any point), so you get parallel lines. For intuition, think about how in painting/photography parallel lines (like railroad tracks) meet in the distance (at infinity) due to perspective projection. If you parallel-transport the eye/camera (no rotation), the point at infinity appears stationary in the image (conversely, if you parallel-transport a line, it will always hit the same vanishing point).

Maybe the author just wanted to point out that in a more complicated situation where the relationships are less obvious — e.g. a plane arbitrarily oriented in 3D space — you can find the shortest path if you compute the normal vector, use it to project a ray from a point, and then find the intersection of that ray and the plane.

• Good answer but how does the last example, relate to the initial idea? I don't see any points and infinities in it. I could put the light source at the foot perpendicular in the plane Commented Mar 29, 2021 at 8:32
• @Buraian - that example is not about distances, it was just my attempt to help you build some intuition regarding the remark immediately preceding it - that "aiming" at the same point at infinity (projecting a ray towards it) from any point produces parallel rays (and a camera lets you "aim"). "I don't see any points and infinities in it." - if you extend the railroad tracks "to infinity" they will meet (and determine) a particular point at infinity (and in a camera, they will actually look like they meet). 1/2 Commented Mar 29, 2021 at 9:28
• @Buraian - The line/plane from the handout is not present in this analogy; it can be introduced as a wall perpendicular to the railroad tracks. Then the shortest distance from any light source to the wall is a ray from the source towards the point at infinity defined by the extension of the tracks. 2/2 Commented Mar 29, 2021 at 9:28

The key is the finite extend of the plane, which is present in most experiments:

Suppose we have a square of length $$a$$ and a point $$q$$, which sits normal to the surface of the square at a distance $$d$$.

If we increase the distance $$d$$ such that $$a \ll d$$ the quoted statement is correct. Hence, if we increase the size $$a$$ of the plane, the distance $$d$$ must increase as well. As stated in your quote: If we think of this in terms of geometrical optics the rays travel (approximately) normal to the surface of the square. Hence, they reach the square (approx.) at the same time. The quality of this approximation only depends on our choice of the condition $$a \ll d$$.

In optics we often encounter the following: We start with a spherical wave $$e^{ikr}/|r|$$, where $$r=(x,y,z)$$. If we place a detector or a lens at a "far distance" $$1\ll z$$, we capture only a "small part" of the spherical wave. Within this "small part" the curvature of the sphere is negligible. Thus, we are allowed to describe the incoming wave as a plane wave $$e^{ikz}$$.