3dgeometrylinear algebralinear-transformationsvector-spaces
Assume I have 3 original points in a 3D object (in 3D space) as A1=<xa,ya,za>, A2=<xa,ya,za>, and A3=<xa,ya,za>. The 3D object is moved and rotated in the 3D space, and the destination points in that object become B1=<xb,yb,zb>, B2=<xb,yb,zb>, and B3=<xb,yb,zb>.
How can I compute the transformation matrix and following that, the degree of rotation in each axis? Basically, I need a matrix that if applied to all points, I get the displaced object.
Concentrating on just the first octant, we can see that a rotation by $\frac{2\pi}{3}$ (counterclockwise of course, with respect to the plane through the tips of $i,j,k$ (this is normal to $(1,1,1)$) sends $i$ to $j$, $j$ to $k$ and $k$ to $i$, where $i,j,k$ are my names for the standard basis. This plane cuts the first octant at an equilateral triangle, and the rotation is just permuting the vertices.
Writing this as a matrix on the left of column vectors, we would have then
$$T=\begin{bmatrix}0&0&1 \\1&0&0\\0&1&0\end{bmatrix}$$
As you can see:
$$T\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix}$$
$$T\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}$$
$$T\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}$$
So it is indeed the transformation that permutes $i,j,k$ cyclically.
This is indeed undefined when you are missing the depth information.
If the rotation axis is fixed, all points are moving on ellipses of various extents, as seen on these cylinders, so the new positions aren't completely undefined.
If you have two pictures of the same object, stereoscopy allows you to reconstruct the 3D model.
Best Answer
Let us call the transformation matrix $\mathbf{M}$, $$\mathbf{M} = \left [ \begin{matrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{32} & m_{32} & m_{33} \end{matrix} \right ]$$ your three original vectors $$\begin{array}{l} \vec{p}_1 = ( x_1 , y_1 , z_1 ) \\ \vec{p}_2 = ( x_2 , y_2 , z_2 ) \\ \vec{p}_3 = ( x_3 , y_3 , z_3 ) \end{array}$$ and their corresponding transformed vectors $$\begin{array}{l} \vec{p}_4 = ( x_4 , y_4 , z_4 ) \\ \vec{p}_5 = ( x_5 , y_5 , z_5 ) \\ \vec{p}_6 = ( x_6 , y_6 , z_6 ) \end{array}$$ so that $$\begin{cases} \mathbf{M} \vec{p}_1 = \vec{p}_4 \\ \mathbf{M} \vec{p}_2 = \vec{p}_5 \\ \mathbf{M} \vec{p}_3 = \vec{p}_6 \end{cases}$$ Let $d$ be the triple product of the first three vectors, $$d = \vec{p}_1 \cdot \left ( \vec{p}_2 \times \vec{p}_3 \right ) = x_1 ( y_2 z_3 - y_3 z_2 ) + x_2 ( y_3 z_1 - y_1 z_3 ) + x_3 ( y_1 z_2 - y_2 z_1 )$$ If $d \ne 0$, there exists a solution: $$\begin{cases} m_{11} = \left ( x_4 ( y_2 z_3 - z_2 y_3 ) + x_5 ( z_1 y_3 - y_1 z_3 ) + x_6 ( y_1 z_2 - z_1 y_2 ) \right ) / d \\ m_{12} = \left ( x_4 ( z_2 x_3 - x_2 z_3 ) + x_5 ( x_1 z_3 - z_1 x_3 ) + x_6 ( z_1 x_2 - x_1 z_2 ) \right ) / d \\ m_{13} = \left ( x_4 ( x_2 y_3 - y_2 x_3 ) + x_5 ( y_1 x_3 - x_1 y_3 ) + x_6 ( x_1 y_2 - y_1 x_2 ) \right ) / d \\ m_{21} = \left ( y_4 ( y_2 z_3 - z_2 y_3 ) + y_5 ( z_1 y_3 - y_1 z_3 ) + y_6 ( y_1 z_2 - z_1 y_2 ) \right ) / d \\ m_{22} = \left ( y_4 ( z_2 x_3 - x_2 z_3 ) + y_5 ( x_1 z_3 - z_1 x_3 ) + y_6 ( z_1 x_2 - x_1 z_2 ) \right ) / d \\ m_{23} = \left ( y_4 ( x_2 y_3 - y_2 x_3 ) + y_5 ( y_1 x_3 - x_1 y_3 ) + y_6 ( x_1 y_2 - y_1 x_2 ) \right ) / d \\ m_{31} = \left ( z_4 ( y_2 z_3 - z_2 y_3 ) + z_5 ( z_1 y_3 - y_1 z_3 ) + z_6 ( y_1 z_2 - z_1 y_2 ) \right ) / d \\ m_{32} = \left ( z_4 ( z_2 x_3 - x_2 z_3 ) + z_5 ( x_1 z_3 - z_1 x_3 ) + z_6 ( z_1 x_2 - x_1 z_2 ) \right ) / d \\ m_{33} = \left ( z_4 ( x_2 y_3 - y_2 x_3 ) + z_5 ( y_1 x_3 - x_1 y_3 ) + z_6 ( x_1 y_2 - y_1 x_2 ) \right ) / d \end{cases}$$
If $\mathbf{M}$ is a pure rotation matrix, it will be orthogonal. If we use $\hat{r}_i = ( m_{i 1} ,\, m_{i 2} ,\, m_{i 3} )$ and $\hat{c}_j = ( m_{1 j} ,\, m_{2 j} ,\, m_{3 j} )$, and $\mathbf{M}$ is orthogonal, then $$\begin{array}{l|l|l|l} \hat{r}_1 \cdot \hat{r}_1 = 1 & \hat{c}_1 \cdot \hat{c}_1 = 1 & \hat{r}_1 \times \hat{r}_1 = 0 & \hat{c}_1 \times \hat{c}_1 = 0 \\ \hat{r}_1 \cdot \hat{r}_2 = 0 & \hat{c}_1 \cdot \hat{c}_2 = 0 & \hat{r}_1 \times \hat{r}_2 = \hat{r}_3 & \hat{c}_1 \times \hat{c}_2 = \hat{c}_3 \\ \hat{r}_1 \cdot \hat{r}_3 = 0 & \hat{c}_1 \cdot \hat{c}_3 = 0 & \hat{r}_1 \times \hat{r}_3 = -\hat{r}_2 & \hat{c}_1 \times \hat{c}_3 = -\hat{c}_2 \\ \hline \hat{r}_2 \cdot \hat{r}_1 = 0 & \hat{c}_2 \cdot \hat{c}_1 = 0 & \hat{r}_2 \times \hat{r}_1 = -\hat{r}_3 & \hat{c}_2 \times \hat{c}_1 = -\hat{c}_3 \\ \hat{r}_2 \cdot \hat{r}_2 = 1 & \hat{c}_2 \cdot \hat{c}_2 = 1 & \hat{r}_2 \times \hat{r}_2 = 0 & \hat{c}_2 \times \hat{c}_2 = 0 \\ \hat{r}_2 \cdot \hat{r}_3 = 0 & \hat{c}_2 \cdot \hat{c}_3 = 0 & \hat{r}_2 \times \hat{r}_3 = \hat{r}_1 & \hat{c}_2 \times \hat{c}_3 = \hat{c}_1 \\ \hline \hat{r}_3 \cdot \hat{r}_1 = 0 & \hat{c}_3 \cdot \hat{c}_1 = 0 & \hat{r}_3 \times \hat{r}_1 = \hat{r}_2 & \hat{c}_3 \times \hat{c}_1 = \hat{c}_2 \\ \hat{r}_3 \cdot \hat{r}_2 = 0 & \hat{c}_3 \cdot \hat{c}_2 = 0 & \hat{r}_3 \times \hat{r}_2 = -\hat{r}_1 & \hat{c}_3 \times \hat{c}_2 = -\hat{c}_1 \\ \hat{r}_3 \cdot \hat{r}_3 = 1 & \hat{c}_3 \cdot \hat{c}_3 = 1 & \hat{r}_3 \times \hat{r}_3 = 0 & \hat{c}_3 \times \hat{c}_3 = 0 \end{array}$$
When you are satisfied you have an orthogonal matrix, pick the Euler or Tait-Bryan angles you wish to use, you can extract the angles by inspecting the matrix $\mathbf{M}$.
I personally do not use Euler angles or Tait-Bryan angles, because they are confusing (no single definition; you need to specify the rotation axes and their order, as Euler angles or Tait-Bryan angles refer to a set of definitions, not one definition) and problematic (gimbal lock in particular). I personally use versors and rotation matrices, and occasionally the axis-angle representation and Rodrigues' rotation formula.
One practical way to split a transformation back into a translation and rotation is to define the rotation as rotation around one of the vectors. This yields translation before a rotation, a rotation, and a translation after the rotation.
Let's assume we know three vectors before the transformation, $\vec{a}_1$, $\vec{a}_2$, and $\vec{a}_3$, and the three vectors after the transformation, $\vec{b}_1$, $\vec{b}_2$, $\vec{b}_3$, but not the transformation itself -- not the rotation $\mathbf{R}$ nor the translation after rotation $\vec{t}$: $$\begin{array}{c} \mathbf{R}\vec{a}_1 + \vec{t} = \vec{b}_1 \\ \mathbf{R}\vec{a}_2 + \vec{t} = \vec{b}_2 \\ \mathbf{R}\vec{a}_3 + \vec{t} = \vec{b}_3 \end{array}$$ Since $\mathbf{R}$ describes a pure rotation, it is an orthogonal matrix.
We can split the above into pre- and post-translation thus: $$\begin{array}{c} \mathbf{R}\left(\vec{a}_1 - \vec{t}_1 \right ) + \vec{t}_2 = \vec{b}_1 \\ \mathbf{R}\left(\vec{a}_2 - \vec{t}_1 \right ) + \vec{t}_2 = \vec{b}_2 \\ \mathbf{R}\left(\vec{a}_3 - \vec{t}_1 \right ) + \vec{t}_2 = \vec{b}_3 \end{array}$$ with $\vec{t} = \vec{t}_2 - \mathbf{R} \vec{t}_1$.
Now, if we pick $\vec{t}_1 = \vec{a}_i$ and $\vec{t}_2 = \vec{b}_i$, with $i$ being any of $1$, $2$, or $3$, and solve $\mathbf{R}$ using the other two vector pairs relative to $\vec{a}_i$ and $\vec{b}_i$, we will find a rotation that combined with the specified translation fulfills the requirements. Depending on the vectors and the rotation -- and in practice this is quite sensitive to rounding errors and such --, there may be multiple solutions. However, this should find a valid solution, unless the original vectors are parallel or degenerate.
For simplicity, let's pick $\vec{t}_1 = \vec{a}_1$ and $\vec{t}_2 = \vec{b}_1$. We now have a pure rotation $\mathbf{R}$, such that $$\begin{array}{c} \mathbf{R}\left(\vec{a}_2 - \vec{a}_1\right) = \vec{b}_2 - \vec{b}_1 \\ \mathbf{R}\left(\vec{a}_3 - \vec{a}_1\right) = \vec{b}_3 - \vec{b}_1 \end{array}$$ The simplest way to construct $\mathbf{R}$ is to find the matrix $\mathbf{R}_1$ that rotates the coordinate system to match that used by the $a$ vectors, and $\mathbf{R}_2$ that rotates the coordinate system to match that used by the $b$ vectors. Then, we can express $\mathbf{R}$ as the inverse rotation to $\mathbf{R}_1$ followed by rotation $\mathbf{R}_2$.
Since rotation matrices are orthogonal, their inverse is their transpose, so $$\mathbf{R} = \mathbf{R}_2 \mathbf{R}_1^T$$
We can construct $\mathbf{R}_1$ and $\mathbf{R}_2$ by orthonormalizing the two vectors we have; the third basis vector is then the cross product of the first two.
Let $$\mathbf{R}_1 = \left [ \begin{matrix} \hat{e}_1 & \hat{e}_2 & \hat{e}_1\times\hat{e}_2 \end{matrix} \right ]$$ and $$\mathbf{R}_2 = \left [ \begin{matrix} \hat{e}_4 & \hat{e}_5 & \hat{e}_4\times\hat{e}_5 \end{matrix} \right ]$$ ie. $\hat{e}_i$ are the column vectors (basis unit vectors) in the rotation matrices.
We can use the first leftover vector to define the $x$ axis unit vector, $$\hat{e}_1 = \frac{\vec{a}_2 - \vec{a}_1}{\lVert \vec{a}_2 - \vec{a}_1 \rVert} $$ $$\hat{e}_4 = \frac{\vec{b}_2 - \vec{b}_1}{\lVert \vec{b}_2 - \vec{b}_1 \rVert} $$ and orthonormalize the second vector against the first one, $$\hat{e}_2 = \frac{\vec{a}_3 - \vec{a}_1 - \hat{e}_1 \left ( \hat{e}_1 \cdot (\vec{a}_3 - \vec{a}_1) \right )}{\lVert \vec{a}_3 - \vec{a}_1 - \hat{e}_1 \left ( \hat{e}_1 \cdot (\vec{a}_3 - \vec{a}_1) \right ) \rVert} $$ $$\hat{e}_5 = \frac{\vec{b}_3 - \vec{b}_1 - \hat{e}_4 \left ( \hat{e}_4 \cdot (\vec{b}_3 - \vec{b}_1) \right )}{\lVert \vec{b}_3 - \vec{b}_1 - \hat{e}_4 \left ( \hat{e}_4 \cdot (\vec{b}_3 - \vec{b}_1) \right ) \rVert} $$ As shown in the matrices above, the third basis unit vector is just the cross product of the two others, $$\hat{e}_3 = \hat{e}_1 \times \hat{e}_2$$ $$\hat{e}_6 = \hat{e}_4 \times \hat{e}_5$$
In practice, you'd construct the unit vectors $\hat{e}_i$ first, copying them to the rotation matrices $\mathbf{R}_1$ and $\mathbf{R}_2$, then compute the rotation via matrix multiplication, $\mathbf{R} = \mathbf{R}_2 \mathbf{R}_1^T$, and finally calculate the translation after rotation, $\vec{t} = \vec{b}_1 - \mathbf{R} \vec{a}_1$.