# Undoing a rotation: determination of origin required to rotate a vector B to equal A

1. Jul 24, 2012

### gge

Given two 2D unit vectors (A and B) in different directions and positions in the same horizontal plane, is it possible to determine the origin (point of rotation) required to make vector B = vector A (same position and direction)? All this assuming that vector B is a rotated (only) version of vector A (i.e., there is a solution). I can find the angle between them (of course), but I don't know around what point I should apply it.

The problem lies in the fact that the vectors are now in an arbitrary coordinate system unrelated to that in which they underwent the original rotation. Essentially, I am now attempting to undo the rotation that was applied.

Any help (even a push in the right direction) would be great. Perhaps it's something silly that has me stumped.

Cheers,

2. Jul 24, 2012

### Staff: Mentor

Re: Undoing a rotation: determination of origin required to rotate a vector B to equa

why not the endpoint of each vector?

3. Jul 24, 2012

### gge

Re: Undoing a rotation: determination of origin required to rotate a vector B to equa

Vector A must stay stationary. Vector B must be rotated around a point in the plane that will make it fall directly on top of vector A.

Picture two vectors of unit length lying in a plane in different directions and locations. Where do you place the origin of rotation so that rotating vector B around that point will cause it to fall directly on top of A (assuming there is a solution).

If I understand your response, rotating vector A and B about their endpoints will only rotate the vectors about themselves.

4. Jul 24, 2012

### SteveL27

Re: Undoing a rotation: determination of origin required to rotate a vector B to equa

Do you mean vectors? Or arrows? Because a vector does not have a position in space. A vector is the equivalence class of all arrows having the same direction and magnitude. A pedantic point, but a vital one. The benefit of vectors is that we are free to choose whichever representative of a vector is most convenient for a given problem. A vector has representatives everywhere.

To make two unit vectors coincide, it's only necessary to rotate one of them by the difference of their arguments (angles) when the vectors are expressed in polar notation. Since they are both unit vectors, we now have both vectors with the same length and direction. So they're the same vector.

To make two unit arrows coincide, first rotate one of them as a vector to make their directions coincide; then map the base points any way that's convenient or that makes sense in the context of your application or problem.

Last edited: Jul 24, 2012
5. Jul 26, 2012

### gge

Re: Undoing a rotation: determination of origin required to rotate a vector B to equa

Thank you for your response Steve. I apologize, my terminology was clearly incorrect. I come from a surveying background, and was trying to frame the question in physics terms.

What I actually have are two coordinate pairs—so four points in all—the first two describe a line A and the second two describe another line B.

The coordinates for line B were formed by applying a 2D rotation matrix to the coordinates of line A (no translations were involved), creating a new rotated version of A.

So you would think that I could just find the angle between them and apply the rotation in the opposite sense to bring line B back so the coordinates of A = B. However, somewhere along the way, the coordinates were moved into a coordinate system with a different origin than when they were rotated. I have no idea where the original origin was (that it was rotated about).

So without applying any translation, is it possible to determine around what point to rotate B so B = A.

Thanks!

6. Jul 26, 2012

### HallsofIvy

Staff Emeritus
Re: Undoing a rotation: determination of origin required to rotate a vector B to equa

With two general line segments (assuming they are of the same length), you will need to first translate A so that its endpoint lies on the endpoint of B. Then you have an angle between the two segments and can rotate.

Or if you simply want A rotated so that it lies along the same line as B, exend the two lines until they intersect. The point of intersection will be the point to rotate about. If the two lines are parallel, you cannot "rotate" A to lie along B.

7. Jul 26, 2012