How to rotate Cartesian coordinate system?

In summary, the conversation discusses the topic of rotating a Cartesian coordinate system in order to achieve specific angles between the new and old axes. The participants suggest using Wikipedia and other sources as well as solving a system of equations to determine the necessary matrix elements for the rotation. A helpful link is also provided. Finally, the conversation shifts to a specific problem of rotating a coordinate system about the z-axis to align with a given line, and suggestions such as using a projection are made.
  • #1
leoleo2
5
0
Hello,
I would like to rotate the Cartesian coordinate system ( i=(1,0,0); j=(0,1,0); k=(0,0,1) ) so that angles between new and the old axes be equal to α, β and γ, respectively. Is any simple way similar to the Euler transformations to accomplish that?
 
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  • #2
leoleo2 said:
Hello,
I would like to rotate the Cartesian coordinate system ( i=(1,0,0); j=(0,1,0); k=(0,0,1) ) so that angles between new and the old axes be equal to α, β and γ, respectively. Is any simple way similar to the Euler transformations to accomplish that?
This link should be very helpful for you. :wink:
 
  • #3
Mandelbroth said:
This link should be very helpful for you. :wink:

Wikipedia really? :)) It's not even close to helpful. :)
 
  • #4
leoleo2 said:
Wikipedia really? :)) It's not even close to helpful. :)
Mathematics pages on Wikipedia are almost religiously checked for accuracy because math people are strict in their pursuit of accuracy. We're crazy like that. If you doubt me, try messing up a formula on a random math page and watch it get fixed, unless it's an arguably correct change, within an hour.

It's very helpful. Wikipedia is always the first place I start looking when I have a question.

If you don't like Wikipedia, though, there's always this alternative. :tongue:
 
  • #5
Mandelbroth said:
Mathematics pages on Wikipedia are almost religiously checked for accuracy because math people are strict in their pursuit of accuracy. We're crazy like that. If you doubt me, try messing up a formula on a random math page and watch it get fixed, unless it's an arguably correct change, within an hour.

It's very helpful. Wikipedia is always the first place I start looking when I have a question.

If you don't like Wikipedia, though, there's always this alternative. :tongue:

OK. I believe you that Wikipedia generally is helpful and I use it too. But I can not find any specific idea on Wikipedia for my specific problem.
 
  • #6
You're basically looking for a matrix A such that the vectors Ai, Aj and Ak form a basis for 3D space and are in angles α, β and γ relative to the vectors i, j and k.

You can form a system of equations from which you can solve the elements of matrix A by requiring that:

1. A must be an orthogonal matrix, i.e. vectors Ai, Aj and Ak form an orthonormal set, too, when i, j and k do.

2. The dot products are [itex] {\bf i} \cdot(A {\bf i} )=cos(\alpha)[/itex], [itex] {\bf j} \cdot(A {\bf j} )=cos(\beta)[/itex], [itex] {\bf k} \cdot(A {\bf k} )=cos(\gamma)[/itex]

These conditions are enough to determine the matrix elements.
 
  • #7
Don't give up. I think this fits your bill: http://inside.mines.edu/~gmurray/ArbitraryAxisRotation/
 
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  • #8
hilbert2 said:
You're basically looking for a matrix A such that the vectors Ai, Aj and Ak form a basis for 3D space and are in angles α, β and γ relative to the vectors i, j and k.

You can form a system of equations from which you can solve the elements of matrix A by requiring that:

1. A must be an orthogonal matrix, i.e. vectors Ai, Aj and Ak form an orthonormal set, too, when i, j and k do.

2. The dot products are [itex] {\bf i} \cdot(A {\bf i} )=cos(\alpha)[/itex], [itex] {\bf j} \cdot(A {\bf j} )=cos(\beta)[/itex], [itex] {\bf k} \cdot(A {\bf k} )=cos(\gamma)[/itex]

These conditions are enough to determine the matrix elements.

Thank you very much for you reply. You stated the problem more clearly than I did.

From the initial conditions (2.) I can define immediately three components of new basis vectors ([itex] A{\bf i}_x = acos(\alpha)[/itex], [itex] A{\bf j}_y = acos(\beta)[/itex], [itex] A{\bf k}_z = acos(\gamma)[/itex]).
For the rest of components I can write the orthogonality (using scalar or/and vector products) and the normalization conditions for new basis vectors. However, this gives system of six quadratic equations with six unknowns which is quite ugly to solve generally. Bedsides that, I would get 8 or 16 different solutions of that system and I doubt existence of more than two solutions for the original problem.
 
  • #9
SteamKing said:
Don't give up. I think this fits your bill: http://inside.mines.edu/~gmurray/ArbitraryAxisRotation/

Thank you very much!
The page looks interesting but it's not clear for me how to correlate the Rotation About an Arbitrary Axis to my problem.
 
Last edited by a moderator:
  • #10
Could someone help me figure out how to rotate a coordinate system about the z-axis such that the the line y = mx + c coincides with the x-axis?
Shouldn't a simple projection of all the coordinates i.e xj = xicos(theta), yj =yicos(theta) and zj = zj cos(theta) work?
 

1. How do I rotate a Cartesian coordinate system?

To rotate a Cartesian coordinate system, you can use the following steps:

  • 1. Determine the angle of rotation in radians or degrees.
  • 2. Create a rotation matrix using the angle of rotation.
  • 3. Multiply the rotation matrix by the original coordinates to get the new rotated coordinates.

2. How do I know the direction of rotation for a Cartesian coordinate system?

The direction of rotation for a Cartesian coordinate system is determined by the sign of the angle of rotation. A positive angle of rotation results in a counter-clockwise rotation, while a negative angle of rotation results in a clockwise rotation.

3. Can I rotate a Cartesian coordinate system around a point other than the origin?

Yes, it is possible to rotate a Cartesian coordinate system around a point other than the origin. To do so, you will need to translate the coordinate system so that the point you want to rotate around becomes the origin. Then, you can perform the rotation and translate the system back to its original position.

4. How do I rotate a 3D Cartesian coordinate system?

To rotate a 3D Cartesian coordinate system, you can use the same steps as rotating a 2D system. However, you will need to use a 3x3 rotation matrix instead of a 2x2 matrix. This matrix will account for rotations around the x, y, and z axes.

5. Is there a specific formula for rotating a Cartesian coordinate system?

Yes, there is a formula for rotating a Cartesian coordinate system. The formula is:
x' = xcosθ - ysinθ
y' = xsinθ + ycosθ
Where x and y are the original coordinates, x' and y' are the new rotated coordinates, and θ is the angle of rotation.

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