Rotation of 3 lines in 3D space

Question

I hope you can help me with a mathematical problem I'm trying to solve (for many days) and so far no success.

Given 2 sets P=[p_1,p_2,p_3] and Q=[q_1,q_2,q_3] of 3 infinite lines each in 3D space (each p_i and q_i (i=1,2,3) is a 3D line represented by a 3D point and a unit vector). so overall we have 6 lines.
I need to find a rotation matrix R (3x3) so that if I rotate all the lines in P by R then I get new lines P_rotate=[p_1_rotate, p_2_rotate, p_3_rotate] so that p_i_rotate and q_i (i=1,2,3) intersect each other.
I know there might not be such a rotation matrix and in some cases there might be infinite rotation matrices (for example if all the 6 lines are on the same plane and the axis of rotation is perpendicular to that plane). but the assumption here that for sure there exists such a rotation matrix and if there are more than one, then I can choose any of them.

This is what I tried to do so far:
It's actually solving 3 equations of 3 variables (the variables are the 3 angles around the axises x,y,z). but the equations were extremely complex that I couldn't solve them (and I'm not even sure they are solvable). You can calculate the rotated lines as a function of the 3 angles. meaning each rotated line will contain a new point and a new vector both as a function of the 3 angles. then I calculate for each line the distance to the relevant line in Q. and I need to find for which angles the 3 distances are zero. I assumed here that first I rotate around Z axis, then around Y axis and then around X axis. I think the order doesn't matter since whatever order I choose I may get different angles but the final rotation matrix will be the same. Anyway, the final 3 equations were extremely long and with many multiplications of the unknown variables so it was actually impossible for me to solve them.

I also tried to solve it in a different way. not using rotation matrix. but Rodrigues' rotation formula. in Rodrigues' formula I rotate around an axis line that passes through the origin by an angle theta. so I actually need to find v_x,v_y,theta where v_x, v_y are the x and y components of a unit vector of the axis line (v_z can be calculated by v_x and v_y since the vector is unit). and theta is an angle of rotation around this axis. again, I got 3 very long and complex equations that I couldn't solve.

Anyone has any idea of how to solve it? or can refer me to a link that (hopefully) explains about it? Will appreciate any help!

Thank you
David

Answer 1

This is not an answer to the question as posed, but rather a rough sketch if the rotations could be about any axis.

Fix the set $P=\{P_1,P_2,P_3\}$ of lines. The set $Q=\{Q_1,Q_2,Q_3\}$ will be moved. Rotate $Q$ about an axis through the origin, with the axis not parallel to the directions of the three $Q$-lines, and not orthogonal to the directions of the three $P$-lines, until $Q_1$ bangs into $P_1$ at point $a$. I believe if the axis is generic enough, $Q_1$ will hit $P_1$. Not sure I have the genericity conditions correct, but some assumptions will work.

Now spin $Q$ about the vector from the origin to $a$ until $Q_2$ bangs into $P_2$ at point $b$. Again I need to assume no accidental degeneracies to ensure the collision. Note the $Q_1 \cap P_1$ intersection is maintained.

Finally, spin $Q$ about the line containing $ab$. This axis is (in general) not through the origin. Barring degeneracies, $Q_3$ will bang into $P_3$, while the $Q_1 \cap P_1$ and $Q_2 \cap P_2$ intersections remain.

Answer 2

As you said, each line can be written, in parametric form, as $$ \left( {\matrix{ x \cr y \cr x \cr } } \right) = \left( {\matrix{ {a_x } \cr {a_y } \cr {a_z } \cr } } \right) + \lambda \left( {\matrix{ {v_x } \cr {v_y } \cr {v_z } \cr } } \right) $$

So we can represent each group of three lines in matricial notation as $$ {\bf X} = \left( {\matrix{ {x_{\,1} } & {x_{\,2} } & {x_{\,3} } \cr {y_{\,1} } & {y_{\,2} } & {y_{\,3} } \cr {z_{\,1} } & {z_{\,2} } & {z_{\,3} } \cr } } \right) = {\bf A} + {\bf V}\;{\bf \Lambda }\quad \quad {\bf X}' = \left( {\matrix{ {x'_{\,1} } & {x'_{\,2} } & {x'_{\,3} } \cr {y'_{\,1} } & {y'_{\,2} } & {y'_{\,3} } \cr {z'_{\,1} } & {z'_{\,2} } & {z'_{\,3} } \cr } } \right) = {\bf A}' + {\bf V}'\;{\bf \Lambda }' $$ where is clear the meaning of the matrices , and in particular that the ${\bf \Lambda }$ and ${\bf \Lambda }'$ matrices are diagonal.

Now if each line of the first group was to intersect the corresponding line in the second group, that means that there are specific values of $\lambda_k$ and ${\lambda '}_j$ such that ${\bf X} ={\bf X}' $, i.e. $$ {\bf A} + {\bf V}\;{\bf \Lambda } = {\bf A}' + {\bf V}'\;{\bf \Lambda }' $$ or $$ \;{\bf \Lambda } = {\bf V}^{\, - \,{\bf 1}} \left( {{\bf A}' - {\bf A}} \right) + {\bf V}^{\, - \,{\bf 1}} {\bf V}'\;{\bf \Lambda }' = diag $$

That translates into a linear system of six equations in the three unknowns $\lambda'$, which then determine the $\lambda$.

If instead each line of the first group was to intercept one line of the 2nd group, different but not necessarily the corresponding one, then we shall have ${\bf X} \, {\bf P} ={\bf X}' $, or v.v, where ${\bf P}$ is one of the $6$ Permutation matrices.

Also, if each line of the first group was to intercept one line of the 2nd group, not necessarily distinct, then we shall have ${\bf X} ={\bf X}' \, {\bf Q} $, where ${\bf Q}$ is one of the $9$ "$0/1$" matrices, having only one $1$ in each column.

Finally, as per your question, we can introduce a Rotation matrix, depending on $3$ angles, which goes to multiply ${\bf X}$ $$ {\bf R}\,\left( {{\bf A} + \,{\bf V}\;{\bf \Lambda }} \right) = \left( {{\bf A}' + {\bf V}'\;{\bf \Lambda }'} \right)\;\left( {{\bf P}\,{\rm or}\;{\bf Q}} \right) $$ to solve by imposing the diagonalization condition as above. i.e. $$ {\bf \Lambda } = {\bf V}^{\, - \,{\bf 1}} \;{\bf R}^{\, - \,{\bf 1}} \;\left( {{\bf A}'{\bf Q} - {\bf R}\,{\bf A}} \right) + {\bf V}^{\, - \,{\bf 1}} \;{\bf R}^{\, - \,{\bf 1}} {\bf V}'\;{\bf \Lambda }'{\bf Q}\; $$ By imposing that the "6" off-diagonal elements of the RHS matrix be null, we get system of $6$ equations in the $3$ unknowns ${\lambda}'$, and in the $3$ rotation angles, for each of the $9$ ${\bf Q}$ matrices we are willing to consider as "crossing conditions".
( let comprise in ${\bf Q}$ also the particular cases of unit and permutation matrices).

Now, the homogeneous system is linear in the ${\lambda}'$ but contains product of $\sin $ and $ \cos$ of the angles, and you are asking for a way to simplify that.

The first way I can see at the moment is that, restarting from the identity $$ {\bf R}\,\left( {{\bf A} + \,{\bf V}\;{\bf \Lambda }} \right) = \left( {{\bf A}' + {\bf V}'\;{\bf \Lambda }'} \right)\;{\bf Q} $$ taking the transposed (indicated by the hat) $$ \left( {\overline {\bf A} + \;{\bf \Lambda }\,\overline {\bf V} } \right){\bf R}^{\, - \,{\bf 1}} = \overline {\bf Q} \left( {\overline {{\bf A}'} + {\bf \Lambda }'\;\overline {{\bf V}'} } \right)\; $$ we can get rid of the rotation, by multiplying the first by the second on the left

$$ \left( {\overline {\bf A} + \;{\bf \Lambda }\,\overline {\bf V} } \right)\left( {{\bf A} + \,{\bf V}\;{\bf \Lambda }} \right) = \overline {\bf Q} \left( {\overline {{\bf A}'} + {\bf \Lambda }'\;\overline {{\bf V}'} } \right)\left( {{\bf A}' + {\bf V}'\;{\bf \Lambda }'} \right)\;{\bf Q} $$ however at the cost of introducing an equation between two quadratic forms, symmetric, so $6$ equations.
Yet this equation has the advantage of not requiring $\bf V$ to be invertible.

At this point might be profitable to express each line with the position vector ($\bf a$) normal to the direction vector ($\bf v$) and to normalize the latter.
That will give that $\overline {\bf A} \, {\bf V}$ has the main diagonal null, while for $\overline {\bf V} \, {\bf V}$ it is unitary.
So we might possibly start from equating the diagonals ...

Sorry for not being this actually an answer, but this is the best help I can offer.

--- P.S. -----

As per your last comment, the case of two groups of incident lines introduces many simplifications with respect to the general case of being all skew, and can be approached quite differently. It is worthy that you open another post. In any case it remains to clarify the intersection scheme (e.g. two lines 1st group could intersect the same line in 2nd group, and the third any one of the remaining?)