# Passive and Active Transformations

1. Jun 12, 2015

### Septimra

Alright. I was looking into a 2D rotation matrix and there are two equations: one is through the transformation of the component of p (always with respect to x,y), x,y into x',y' and the other is through the transformation of the unit vectors i,j into i',j'.

In a sense 1 is passive the other is an active transformation. I do understand the difference between the two, but my question is that why do we get the different matrices if when we derive them, we do so with a clockwise orientation?

Hope its clear, If not--I'd be happy to clarify.

2. Jun 12, 2015

### Hawkeye18

Can you clarify. I guess I understand what you mean, but just to be sure.

3. Jun 12, 2015

### Septimra

LOL Okay. I got the answer. Welp it seems like they way you derive the matrix you have to say consistent, otherwise you fall in a a trap.
https://en.wikipedia.org/wiki/Active_and_passive_transformation#/media/File:PassiveActive.JPG
Staring at this image for a bit and making sense of this reasoning: http://math.stackexchange.com/quest...erent-representations-of-2d-rotation-matrices

helped. You need to use a point relative to the rotated basis once you rotate the basis. You cannot use a rotated basis on a point relative to a the old basis, but with rotational matrices you only an inverted answer and nothing any more alarming.

4. Jun 12, 2015

### Fredrik

Staff Emeritus
The convention that I'm familiar with is to rotate vectors counterclockwise. A rotation is linear, so if we know how the basis vectors are rotated, we will know how all vectors are rotated. If you rotate (1,0) by an angle $\theta$ counterclockwise, the result is $(\cos\theta,\sin\theta)$. The same rotation applied to (0,1) yields $(-\sin\theta,\cos\theta)$. It follows immediately from this that the matrix of the rotation operator with respect to the standard ordered basis is
$$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}.$$ The rotated basis vectors end up as the columns of the matrix. If you don't understand this, I recommend the FAQ post that I linked to in the post I linked to above.