Difference between revisions of "Inverse transformation"
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The equation above is now solved for <math>\vec{\mathbf{q}}_1</math>:<br/> | The equation above is now solved for <math>\vec{\mathbf{q}}_1</math>:<br/> | ||
:<math>\begin{align} | :<math>\begin{align} | ||
− | \vec{\mathbf{q}}_1&=\mathbf{R} \vec{\mathbf{q}}_0 + \vec{\mathbf{p}}\\ | + | \vec{\mathbf{q}}_1&=\mathbf{R} \vec{\mathbf{q}}_0 + \vec{\mathbf{p}} & & \\ |
− | \vec{\mathbf{q}}_1-\vec{\mathbf{p}}&=\mathbf{R} \vec{\mathbf{q}}_0\\ | + | \vec{\mathbf{q}}_1-\vec{\mathbf{p}}&=\mathbf{R} \vec{\mathbf{q}}_0 & & \\ |
− | \mathbf{R}^{-1}(\vec{\mathbf{q}}_1-\vec{\mathbf{p}})&=\mathbf{R}^{-1}\mathbf{R} \vec{\mathbf{q}}_0\\ | + | \mathbf{R}^{-1}(\vec{\mathbf{q}}_1-\vec{\mathbf{p}})&=\mathbf{R}^{-1}\mathbf{R} \vec{\mathbf{q}}_0 & & \\ |
− | \mathbf{R}^{-1}\vec{\mathbf{q}}_1-\mathbf{R}^{-1}\vec{\mathbf{p}}&=\vec{\mathbf{q}}_0 \qquad\quad \rightarrow \qquad\quad \vec{\mathbf{q}}_0=\underbrace{\mathbf{R}^{-1}}\vec{\mathbf{q}}_1+(\underbrace{-\mathbf{R}^{-1}\vec{\mathbf{p}}})\\ | + | \mathbf{R}^{-1}\vec{\mathbf{q}}_1-\mathbf{R}^{-1}\vec{\mathbf{p}}&=\vec{\mathbf{q}}_0 \qquad\quad \rightarrow \qquad\quad \vec{\mathbf{q}}_0=&\underbrace{\mathbf{R}^{-1}}&\vec{\mathbf{q}}_1+(\underbrace{-\mathbf{R}^{-1}\vec{\mathbf{p}}})\\ |
+ | & &\mathbf{R}_i& \qquad\qquad \vec{\mathbf{p}}_i\\ | ||
\end{align}</math> | \end{align}</math> | ||
[[Category:Article]] | [[Category:Article]] | ||
[[Category:Transformations]] | [[Category:Transformations]] |
Revision as of 16:25, 17 June 2014
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Let be a general homogeneous transformation matrix. The inverse transformation corresponds to the transformation that reverts the rotation and translation effected by . If a vector is pre-multiplied by and subsequently pre-multiplied by , this results in the original coordinates because and multiplication with the identity matrix does not change anything (see transformations).
The general homogeneous transformation matrix for three-dimensional space consists of a 3-by-3 rotation matrix and a 3-by-1 translation vector combined with the last row of the identity matrix:
As stated in the article about homogeneous coordinates, multiplication with is equivalent in cartesian coordinates to applying the rotation matrix first and then translating the coordinates by :
The equation above is now solved for :