# How can I calculate the transformation of two 3D triangles?

Given two triangle I have the transformation (three rotation followed by three translation)of both the triangles. How can I calculate the transformation between two triangles? A numerical example will help. Thanks in advance!

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I don't understand your parenthetical remarks. Is your question "How do calculate a linear transformation of one given triangle into another?" If so, what is the rotation and translation part of your question? Also, if that is your question, you can calculate it by change of basis formulas. –  BBischof May 1 '11 at 15:45
Suppose I have one triangle ABC and another one DEF. ABC has the following transformation(three rotation in radians followed by three translation): 45.1983,-87.0796,44.5519,-7.279,-58.82,-89.988 And DEF has transformation as follows: 88.1941,-21.0735, 53.7017,-24.064,-67.799,-77.009. How to calculate the transformation between these two triangles? –  Irina May 1 '11 at 20:24
What do you mean the triangle has transformations? I thought maybe you meant symmetries, but I don't see how a translation will give a symmetry, so I am clueless. –  BBischof May 1 '11 at 20:34
I defined a news coordinate system such that point A becomes the origin, x-coordinate is the vector AB, z-coordinate is the cross product of x-coordinate and vector AC, and y -coordinate is the cross product of x and y coordinate. This way you can get a new transformation for the triangle ABC. I defined the same thing for DEF and now I want to superimpose DEF on ABC. How can I calculate the transformation between these two triangles? –  Irina May 1 '11 at 21:12
I am so sorry but I am still confused as to what you are asking. I am ok with the definition of your triangle ABC. But I still don't know what you mean by transformation. –  BBischof May 2 '11 at 4:55

Given that this is true, if the matrices for triangle $ABC$ are $R_1,R_2,R_3$ and the translation is $T_1$, and for $DEF$ are $R_4,R_5,R_6,T_2$, and $A'B'C'$ is triangle $ABC$ in standard position (after the rotation/translation) you have $A'B'C'=T_1+R_3R_2R_1ABC=D'E'F'=T_2+R_6R_5R_4DEF$. Then $DEF=R_4^{-1}R_5^{-1}R_6^{-1}(T_1+R_3R_2R_1ABC-T_2)$
If you translate it along $x$, you don't need radians-that is an angle. If you do your rotations around fixed space directions, I believe there are positions you cannot achieve, but am not sure. –  Ross Millikan May 3 '11 at 2:04