Why Do Different Definitions of Rotation Matrices Exist in Mathematics?

[LagrangeEuler]

Happy new year. Why everybody uses this definition of rotation matrix$$R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\[0.3em] \sin\theta & \cos\theta \\[0.3em] \end{bmatrix}$$
? This is clockwise rotation. And we always use counter clockwise in mathematics as a positive direction
$$R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\[0.3em] -\sin\theta & \cos\theta \\[0.3em] \end{bmatrix}$$

 

[S.G. Janssens]

I'm not sure you are correct. Look at what is written here about rotations in 2D, in the section "In two dimensions". In the standard right-handed system, your first matrix represents counterclockwise rotations for positive $\theta$ and clockwise rotations for negative $\theta$.

 

[LagrangeEuler]

I think I am. Please look here

 

[S.G. Janssens]

Aha, I think we are both right, but I was tacitly assuming you were talking about active transformations, while from the video it is apparent that you were in fact talking about passive transformations. See the discussion on the same wiki page under the section "Ambiguities".

 

[LagrangeEuler]

Ok. So first variant
$$R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\[0.3em] \sin\theta & \cos\theta \\[0.3em] \end{bmatrix}$$
is when I use rotation of vector in the system. And second is when I rotate system and vector stays fixed
$$R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\[0.3em] -\sin\theta & \cos\theta \\[0.3em] \end{bmatrix}$$
Thanks!

 

[FactChecker]

Yes. You need to carefully distinguish between the rotation of the coordinate system from one orientation to another, versus the rotation of a structure in a fixed coordinate system. You need to pick the appropriate transformation for each.

 

[rs1n]

Yes. You need to carefully distinguish between the rotation of the coordinate system from one orientation to another, versus the rotation of a structure in a fixed coordinate system. You need to pick the appropriate transformation for each.

Rotating the object by an angle of $\theta$ is the same as rotating the coordinate system by $-\theta$ -- it's all relative. For all intents and purposes, you can just consider only rotating the object (and changing the angle to its negation where needed).

Ok. So first variant
$$R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\[0.3em] \sin\theta & \cos\theta \\[0.3em] \end{bmatrix}$$
is when I use rotation of vector in the system. And second is when I rotate system and vector stays fixed
$$R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\[0.3em] -\sin\theta & \cos\theta \\[0.3em] \end{bmatrix}$$
Thanks!

The second version is basically the same as the first version except if your replace $\theta$ with $-\theta$ and use the fact that cosine is an even function whereas sine is an odd function.