Rotation of Point in 3D Space

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Hello Friends,
I have one problem related to rotation of point in 3D Space.
Suppose I have one point with X, Y and Z Co-Ordinate.
And now I want to rotate it by user defined degree (e.g 30 Degree) in 3D Space by User Define Axis of Rotation Axis (it may be x-axis, y-axis or z-axis) and Most important Must be rotated by User Define Given point with X, Y and Z Co-Ordinate.

In Short,
Rotate Given Point (with X, Y and Z Co-Ordinate)
1. By User Define Degree
2. By User Define Axis Of Rotation And Most important
3. Around User Define Given Point (with X, Y and Z Co-Ordinate)
I found Good link over here,
I think It Does not Satisfy Third Condition………….
Thaks…all…….

Posted 16-Mar-11 23:10pm

Pritesh Aryan

Updated 16-Mar-11 23:24pm

v4

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2 solutions

Solution 2

I think this totally explains how to solve this simple problem: http://www.euclideanspace.com/maths/algebra/matrix/orthogonal/rotation/index.htm[^].

—SA

Posted 18-Mar-11 9:31am

Sergey Alexandrovich Kryukov

v2

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Espen Harlinn 21-Mar-11 5:06am

Good link, my 5

Sergey Alexandrovich Kryukov 21-Mar-11 5:19am

Thank you.
I think it explains it all.
--SA

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Manfred Rudolf Bihy · Accepted Answer · 2011-03-17T03:10:00

Solution 1

This little elaboration will help you along the way: http://www.euclideanspace.com/maths/geometry/affine/aroundPoint/index.htm[^].
It shows how to combine the rotation matrices with a translation to achieve the rotation about a given point. The page you found gives you the rotation matrices for each axis rotation R₀. The article I pointed to gives you the two transform matrices T^-1 and T.
The resulting matrix for the rotation about the point (x,y,z) is then:
R_(x,y,z) = T^-1 * R₀ * T

Alternative notation:
R_(x,y,z) = T^-1 x R₀ x T

Best Regards,

Posted 17-Mar-11 3:10am

Manfred Rudolf Bihy

Updated 18-Mar-11 3:13am

v7

Comments

Pritesh Aryan 17-Mar-11 9:54am

Thank you so much for answering.......now trying solution given by you........thanks...........

Manfred Rudolf Bihy 17-Mar-11 10:33am

Great, I hope it works for you!

Espen Harlinn 17-Mar-11 10:14am

Nice - even properly formatted, good effort - 5!

Manfred Rudolf Bihy 17-Mar-11 10:35am

Just a couple of <sup> & <sub> tags, but what an effect! :)
Thanks for your support, Espen. It has not gone unnoticed.

Sergey Alexandrovich Kryukov 17-Mar-11 15:05pm

Yes, this is how I write formulas, plus some Unicode characters. Your '*' is not quite right. So format can be improved :-)
--SA

Manfred Rudolf Bihy 18-Mar-11 9:12am

Thanks SA for pointing that out. When I did my studies in CS I think it was more like an x for matrix multiplication. Correct me if I'm wrong. :)

Sergey Alexandrovich Kryukov 18-Mar-11 15:37pm

You're right. Concrete math notation does not matter much, I only say you generally need Unicode characters to show the operation. '*' is just for code, in mathematics used for complex conjugation, for example.
--SA

Sergey Alexandrovich Kryukov 17-Mar-11 15:04pm

Manfred, I have a doubt here (need to check up, but one thing I see immediately).
Your solution does not always exist, because not every matrix is invertable (this is like division by 0), but the rotation is always solvable. This is a contradiction. The solution should be always valid. So your answer cannot be correct by that reason.

Actual solution is simple and should involve only one matrix, I thing.

Don't you think?

--SA

Manfred Rudolf Bihy 18-Mar-11 9:11am

Just look at the transformation matrix. There is no linear dependency because it is already a unity transform with additional parameters in one column. I owe you the proove for that, but from all that I know about matices and linear dependencies thereof it should be impossible not to invert them. After all a transformation in space (3D) is always invertible. Move from A -> B and then move from B -> A, enough said I think.
The information of the emtpy translation is the unity matrix BTW. So not moving at all is also always invertible. :)

If you still have doubts I think I could elaborate on this theme. (It might involve some studying though)

Sergey Alexandrovich Kryukov 18-Mar-11 15:30pm

I just shoot off the heap on the formal basic. The task is actually way too simple. Rotation is just one matrix.
--SA

Sergey Alexandrovich Kryukov 18-Mar-11 15:34pm

The thing is that the rotation matrix is not your general transformation matrix, so rotation is always invertable. I even found C# code testing a matrix for being a rotating one.
Actually, I posted a reference in my Answer, think it will explain everything - please see.
--SA