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Find coordonates of a point relative to a second plane in 3D

by dot_binary
Tags: coordonates, plane, point, relative
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dot_binary
#1
Nov10-12, 05:30 AM
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In a 3D plane, there is a point with it's X,Y and Z coordonates known and a second 3D plane within the first plane with it's origins X,Y and Z coordonates relative to the parent plane origin.

How do I find the X,Y and Z coordonates of that point relative to the origin of the second 3D plane taken in consideration that the second one may rotate on all axis and change position.
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#2
Nov10-12, 08:29 AM
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"Change position", I take it, means to translate the origin. If the new coordinate sytem has center at [itex](x_0, y_0, z_0)[/itex] in the old coordinates then a point that has coordinates (x, y, z) in the old coordinate system will have coordinates [itex](x- x_0, y- y_0, z-z_0)[/itex] in the new coordinate system.

"Rotate on all axis" is harder. A rotation about any axis can be reduced to a series of rotations about the coordinate axes and each of those can be written as a matrix product.

Rotation about the z-axis through angle [itex]\theta[/itex] is given by
[tex]\begin{bmatrix}cos(\theta) & -sin(\theta) & 0 \\ sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]

Rotation about the yaxis through angle [itex]\theta[/itex] is given by
[tex]\begin{bmatrix}cos(\theta) & 0 &-sin(\theta) \\ 0 & 1 & 0 \\ sin(\theta) & 0 & cos(\theta) \end{bmatrix}[/tex]

Rotation about the x-axis through angle [itex]\theta[/itex] is given by
[tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & cos(\theta) & -sin(\theta)\\ 0 & sin(\theta) & cos(\theta)\end{bmatrix}[/tex]


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