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3D Rotation Matrix

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RG on 24 Jul 2017
Commented: Alec on 23 Oct 2019
Hi all, I'm new here and am trying to obtain a 3D rotation matrix.
Currently, I have a set of n points: (x1,y1,z1), (x2,y2,z2)...(xn,yn,zn)
from which I have designated one particular point as the origin, and obtained an x'- and y'- vector, based on some features. It is unimportant, for this question, how I chose my origin and got the x'- and y' vector. From the x'- and y' vector, I can also obtain the z'-vector by taking cross product of x' and y'.
Now, I would like to rotate all these points about the origin I have defined, such that the x'-, y'- and z'- vectors are in line with the x-, y- and z- axes respectively.
Hence, I am looking for a 3x3 rotation matrix, R, that can be applied to all points such that:
Transformed (x,y,z) = R * Original (x,y,z)
An example of the vectors to be mapped would be say:
x': [-0.2831, -0.9246, 0.2548]' to be mapped to x: [1 0 0]
y': [0.9242, -0.1919, 0.3303]' to be mapped to y: [0 1 0]
z' will be mapped automatically if the above 2 are satisfied.
Thank you for your help!

  1 Comment

Walter Roberson
Walter Roberson on 24 Jul 2017
Duplicates but this one appears to have been updated so I have closed that.

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Accepted Answer

Honglei Chen
Honglei Chen on 24 Jul 2017
I'm not sure if I fully follow your question, but here is my understanding. The first step is to find the coordinates of all points in the new coordinate system. For that you'll have to first subtract origin from all points and then convert all points to the local coordinate system. That conversion matrix should just be R1 = [x' y' z']'. If you have access to Phased Array System Toolbox, you can consider using the following function
You can then rotate the points in the new coordinate system using Euler matrices. If you want again you can use the following functions:


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More Answers (1)

Herve Hugonnet
Herve Hugonnet on 20 Jun 2019
The simplest if you do not have the phase array system toolbox is to define anonymous functions :
rotx = @(t) [1 0 0; 0 cos(t) -sin(t) ; 0 sin(t) cos(t)] ;
roty = @(t) [cos(t) 0 sin(t) ; 0 1 0 ; -sin(t) 0 cos(t)] ;
rotz = @(t) [cos(t) -sin(t) 0 ; sin(t) cos(t) 0 ; 0 0 1] ;

  1 Comment

Alec on 23 Oct 2019
Thanks for this

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