Let the eye be at

, and the lines be on the

plane and parallel to the

axis, each at distance s from the

-axis. Let the eye be facing towards the positive

-axis. Let the plane

be at distance

from the eye. Let each point on that plane have coordinates

(they are analogous to

actually).

Here is a more informal derivation. The lines intersect the plane

at

and

. Since the vanishing point is directly straight ahead, the projections of these lines on the plane

must intersect at the point

. Since the lines are on a plane and we are projecting them on another plane, their projections must be lines also. Thus we have that the projections of these lines on plane

are the lines starting at

,

and intersecting at

in

coordinates. From these you can use simple trig to deduce that the angle between these two projections must be

on the plane.

Surprisingly, this angle does not depend on the distance we have chosen for the plane. It seems counter-intuitive.

Here is the more complicated derivation that uses some vectors. The idea is to write down an equation of the line going from the eye at

to a point on one of the lines

. Then we need to find the intersection of this line with the plane

. This will be at coordinates

.

To get the equation of a line, we need a position vector

and a direction vector

, which we'll multiply by a parameter

to get the vector equation for the line to be

. So, choose the position vector

and the direction vector be from that point to the eye. By the rules of vector addition, this is

. Thus the vector equation for the line is

Since we want the projection onto the plane

, we need to have the x-coordinates match, which means

. From this we get that

and

. Subbing both of those in the equation of the line, we see that the intersection point is

Now, if we set

, that is we are looking at the intersection between the line and the plane

, we will simply get

. However, the interesting part is when we let it go to infinity.

This is the same result as before! Only this time we have it more rigorously. Furthermore, since the coordinates of the projection of the line on the plane

are

, then in

coordinates, this is just

. These are just parametric equations of the line in parameter

. They form a system of parametric equations

From these we get that

and

, which is the equation of the projection of the line onto the plane

. You can see that it is a line! the other line has equation

. With trig you can get that the angle between one line and the vertical axis is

and the angle between the lines is just

Here is the reason why I think this doesn't depend on the distance

of the plane

. Say we are looking at two parts of the line. One at distance

and the other at distance

. The projections on the plane

will BOTH involve the parameter

, but since we are looking at the relation between those two distances

and

, we will have a sort of cancellation of the parameter

. I think this is a neat result.