# calculate the new position after rotation

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hello everyone.

if you know a 3D points coordinates, and then rotate that point around the point(0,0,0)(so that the distance to the center point stays the same), how do you then calculate the new coordinates?

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hello everyone.

if you know a 3D points coordinates, and then rotate that point around the point(0,0,0)(so that the distance to the center point stays the same), how do you then calculate the new coordinates?

The new coordinates should be the result of whatever rotation method you are using. How are you performing the rotation?

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the rotation is around the Z axis, the Y axis, and the X axis. I hope this is clear enough.

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the rotation is around the Z axis, the Y axis, and the X axis. I hope this is clear enough.

The question is how the rotation is represented. The typical representation is a matrix, so that the rotation is performed as a vector by matrix multiplication resulting in a another vector. It is also possible to use a representation as quaternion directly. Other representations need a conversion first (like an axis-angle pair, AFAIK).

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i am not good at this kind of math, but i guess i am using Euler Angles.

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i am not good at this kind of math, but i guess i am using Euler Angles.

Euler Angles is a representation rotations where 3 angles are given and the order of 3 specific axes of rotation is defined. This is not well suited for application. Hence it is usual to convert each particular of the 3 axis/angle pairs of the Euler Angles to the corresponding matrix representation and multiply all 3 matrices to yield in the combined rotation. Then multiply the position vector and the rotation matrix as already mentioned.

The are some caveats when doing so: E.g. you have to define whether positive angles mean to rotate CW or CCW w.r.t. the axis of rotation, and you have to define whether you use row or column vectors, because that has an influence on the matrix layout as well as on the order of matrices when multiplying them. Look e.g. this section on wikipedia.

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thanks, that works

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