3dtransformationtriangles
Suppose I am given a triangle ABC and its corresponding vertex coordinates in 3D. I want to transform ABC in such a way so that vertex A lies on global (0,0,0) coordinate, B lies on (dist, 0, 0) where dist is the distance between A to B before transformation and C lies on the xy plane.
How can I calculate the transformation for the above case? What would be the rotation and the translation component for such transformation?
The columns $\mathbf h_i$ of a transformation matrix $H$ are the images of the basis vectors expressed in coordinates relative to the “output” basis. If Hpw is an affine transformation matrix that converts from local to global coordinates, then its first three columns are the directions of the local coordinate axes expressed in global coordinates, while the last column is the global homogeneous coordinate vector of the local frame’s origin.† There’s no need to invert anything in order to plot these coordinate axes: just draw an arrow starting at $\mathbf h_4$ and ending at $\mathbf h_4+k\mathbf h_i$, where $k$ is the desired scale factor. Incidentally, if the linear part of $M$ is a rotation matrix $R_{3\times3}$, then since $R_{3\times3}^{-1}=R_{3\times3}^T$, its rows are the local (plane) coordinates of the world coordinate axis directions.
I’m at a bit of a loss as to why, if you indeed used the columns of inv(Hpw) to generate your plot, it looks approximately correct. Since the linear part of your matrix is a rotation, the columns of its inverse are its rows. The second and third rows in particular point in radically different directions than the second and third columns and would result in a rather different picture than the one you’ve got in your question:
I suspect that you didn’t actually use the columns of inv(Hpw) to produce that plot, or that there are mutually-canceling errors. Rotations and the space $\mathbb R^3$ have several convenient properties that can combine to produce numerically-correct results from mathematically incorrect computations.
†: Don’t ignore the last row: The $\mathbf h_i$ are homogeneous coordinate vectors. The zeros at the bottom of the first three columns indicate that the images of the corresponding basis vectors, which are points at infinity, are also points at infinity; the last column is a finite point, which is what one would like for the local coordinate system origin. If $H$ represents a projective transformation, it might not have those zeros in the last row: images of the basis points at infinity can be finite—the vanishing points of lines parallel to those coordinate axes.
Best Answer
First, subtract the coordinates of A from those of all three vertices, giving the translation (assuming translation is done before rotation -- they don't commute).
This gives vertices (0,0,0), (bx,by,bz) and (cx,cy,cz) as vertices.
Define the vector
u = (bx,by,bz)
pointing from A to B, and normalise it to the unit vector
U = ( 1 /|u|) u = ( 1 / √( bx^2 + by^2 + bz^2 ) ) (bx,by,bz)
Take the cross product
w = u × (cx,cy,cz),
which is at right angles to the triangle. Normalise it to give the unit vector
W = ( 1 /|w|) w .
Find the cross-product
V = U ×W ,
automatically a unit vector.
In coordinates corresponding to the basis {U, V, W }, the triangle lies as you want. The rotation matrix that carries the usual (1,0,0) to U, the usual (0,1,0) to V, and the usual (,0,0,1) to W has U, V and W as its columns. You want the other direction, so take the inverse — which for a rotation matrix is just the transpose.
This is not "the transformation for the above case", because you did not specify whether C should lie on the y>0 half plane.