Analytical vector math: Prove or disprove this conjecture.

Question

I'm a game developer trying using a bit of math to optimize some code for a 2 dimensional game. I ran into an interesting situation, and need help in proving this conjecture I came up with.

Suppose this hypothetical situation:

Vector A, a random direction vector shot out from the origin. If I were to take the normal of vector A, I would have vectors N1 and N2 (illustrated in the graphic below, [orange vectors]). Now if I were to take both N1 and N2 and extend them from the center of the circle, one of the vectors will always point to the closest point on A from that circle [green dot]. If I were then to find the unit vector of that normal and take the radius of that circle and multiply those two together [red line], I would get point on the circle [blue dot] closest to the line of A.

Though this situation operates with some nice conditions:

• I have a circle (not ellipse) that is always in quadrant I. The whole circle remains in the quadrant, and is not cut off by any of the axis. This means that the origin will never be to the right of center point of the circle.
• The vector shot out of the origin is always positive. The magnitude makes the vector always extend past circle.
• The vector and the circle never touch each other.

I don't have any idea on how to prove this mathematically, hence why I'm calling this a conjecture for now.

I would greatly appreciate help of any kind.

Answer 1

The conjecture is false as stated. Draw a circle whose center is very near the line spanned by the vector $\mathbf A$ and whose radius is very large such that the circle intersects the line spanned by $\mathbf A$. The minimum distance from the circle to this line is then clearly $0$, but the point your algorithm produces won't lie on the line itself.

However, if the circle does not intersect the line, the point you arrive at will indeed be the point on the circle that is closest to the line spanned by $\mathbf A$. To see that, you only have to realize that this is the point of the circle that points directly to the orthogonal projection of the circle's center onto the line spanned by $\mathbf A$.

For some physical reasoning, imagine that the line spanned by $\mathbf A$ is the surface of the earth, and that you hang a little weight on a string above the surface of the earth. The direction it is going to point in is the direction of gravity, pointing straight down, and that will be closest to the earth.

Answer 2

You can prove the first part of your conjecture using the Pythagorean Theorem. Pick any other point on your line, and make a triangle from the green point, the blue point, and your new point. You now have a right triangle. The blue-green line and the green-new line are perpendicular. $BG^2+GN^2=BN^2$, which implies that any other point is farther from the blue point because $BN>BG$. You can do something similar for the second part by extending a circle around the green point until it hits the circle.