Given $3$ infinite lines $L_1$, $L_2$, $L_3$ in $R^3$ ($3D$ space), each line is represented by two $3D$ points, the line $L_1$ is rotating around the axis line $L_3$ until it intersects with the line $L_2$ (or in other words, until there's a plane that $L_1$ and $L_2$ span). Rotation of a line around axis line means that every point on the line is always in the same distance from the axis line.

I need to find the angle that $L_1$ needs to rotate (around $L_3$) in which it intersects with $L_2$ (or none if the lines don't intersect). Anyone knows how to find it? I've been trying to solve it for quite a while.

A few things I already tried:
The surface of revolution created by rotating the line $L_1$ around $L_3$ is hyperboloid (rotated in space). I was trying to find the equation of the hyperboloid (given $L_1$ and $L_3$) and then to find the intersection point(s) between the hyperboloid and the line $L_2$. But I didn't succeed to find the hyperboloid equation.

Also, I wrote code (in matlab) that rotates a line around another line. so I can have the equation of $L_1$ rotated by the angle theta around $L_3$ (equation depending by one variable - theta ($\theta$)). and I need to find for which $\theta$ there's a point intersection between $L_1$ and $L_2$. The problem is that the equation is extremely long so I couldn't really continue and find the theta in which there's an intersection point.

I'd be happy for any suggestion that can help.
Thank you very much

  • $\begingroup$ This is so sketchy I won't even pretend it's a full answer. But anyway: Maybe you could express the minimum distance between L1 and L2 as a function of the rotation angle about L3. The angle where that minimum distance goes to zero is your angle. $\endgroup$ – bob.sacamento Apr 20 '17 at 15:19
  • $\begingroup$ Choose any plane containing the $L_3$ axis. The rotating $L_1$ intersects the plane in a single point (unless $L_1$ intersects $L_3$ or is parallel to it). During rotation the point draws two branches of a hyperbola. Same result produces $L_2$ rotation w.r.t the same axis. If you manage to find hyperbolas' intersecions, you'll be able to find points on $L_1$ and $L_2$ which need to meet. Then you probably will find angle(s) sought. $\endgroup$ – CiaPan Apr 20 '17 at 15:21
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    $\begingroup$ @David The hypar 1 sheet of revolution with $ (L_1,L_3)$ as generator and axis respectively may not even cut $L_2!$ $\endgroup$ – Narasimham Apr 20 '17 at 17:30
  • $\begingroup$ @Narasimham I know that, and I also wrote in my original question that the lines might not intersect at all. $\endgroup$ – David Apr 22 '17 at 19:31
  • $\begingroup$ @bob.sacamento I tried that (the last paragraph in my question). Didn't succeed that way. $\endgroup$ – David Apr 22 '17 at 19:55

Sketch of a solution: Call the rotation $R$. The point $P$ at which $RL_1$ and $L_2$ intersect (if at all), is the image $RQ$ of a point $Q$ on $L_1$ that is the same distance from the axis of rotation $L_3$ as $P$. This problem is thus equivalent to finding $P\in L_2$ and $Q\in L_1$ that are equidistant from $L_3$ and lie in a plane perpendicular to it. One way to find these two points is to parameterize $L_1$ and $L_2$ and then use well-known formulas for the distance between a point and line in 3-D. Once you have these two points, computing the angle formed by them and the rotation axis is straightforward.

Update: A better approach to finding $P$ and $Q$ is to parameterize $L_3$ as $S(\lambda)=(1-\lambda)A+\lambda B$, where $A$ and $B$ are any two convenient points on this line. You can then find $P(\lambda)$ and $Q(\lambda)$, the intersections of $L_1$ and $L_2$, respectively, with the plane perpendicular to $L_3$ through $S$. Solve $\|P-S\|=\|Q-S\|$, which you can square to get a simple quadratic equation, for $\lambda$, and you have the three points you need to compute the required rotation angle. There may, of course be zero, one or two solutions.

Update (expanding on my comment below): Once you have the points $P$, $Q$ and $S$, an appropriate transformation matrix can be constructed without explicitly computing any angles.

Let $\mathbf u$ be a unit direction vector for $L_3$ ($B-A$ normalized will do), $\mathbf v=(P-S)/\|P-S\|$ and $\mathbf w=(Q-S)/\|Q-S\|$. Since you haven’t specified an orientation for $L_3$, it’s not clear what you mean by a counterclockwise rotation, but let’s say it means that you’d like the rotation angle to be in the range $[0,\pi]$. Examine the sign of the scalar triple product $\mathbf u\cdot\mathbf v\times\mathbf w=\det\begin{bmatrix}\mathbf u&\mathbf v&\mathbf w\end{bmatrix}$. If this is negative, then reverse $\mathbf u$. If it’s zero, then either $L_1$ and $L_2$ already intersect at $P$ or $\mathbf w=-\mathbf v$, in which case the rotation angle will be $\pi$ and there’s no need to make any adjustments. In fact, there’s no need to make any adjustment at all: the orientation changes cancel when the change-of-basis matrices are combined below, so you end up with the same transformation matrix either way.

Now, both $\mathscr B_1=(\mathbf u,\mathbf v,\mathbf u\times\mathbf v)$ and $\mathscr B_2=(\mathbf u,\mathbf w,\mathbf u\times\mathbf w)$ are right-handed orthonormal bases for $\mathbb R^3$. This means that the matrices $$\begin{align}R_1&=\begin{bmatrix}\mathbf u&\mathbf v&\mathbf u\times\mathbf v\end{bmatrix} \\ R_2&=\begin{bmatrix}\mathbf u&\mathbf w&\mathbf u\times\mathbf w\end{bmatrix}\end{align}$$ represent rotations that map the standard basis onto $\mathscr B_1$ and $\mathscr B_2$, respectively. Therefore, the homogeneous transformation matrix that represents the rotation you’re looking for is $$R=\left[\begin{array}{c|c}R_2&S\\\hline0&1\end{array}\right]\left[\begin{array}{c|c}R_1&S\\\hline0&1\end{array}\right]^{-1}.$$ You can avoid the matrix inversion by taking advantage of block structure and the fact that $R_1$ is orthogonal: $$\left[\begin{array}{c|c}R_1&S\\\hline0&1\end{array}\right]^{-1}=\left[\begin{array}{c|c}R_1^{-1}&-R_1^{-1}S\\\hline0&1\end{array}\right]=\left[\begin{array}{c|c}R_1^T&-R_1^TS\\\hline0&1\end{array}\right]$$ and multiplying it all out yields $$R=\left[\begin{array}{c|c}R_2R_1^T&(I-R_2R_1^T)S\\\hline0&1\end{array}\right].$$ Note that you may use any convenient point on the line instead of $S$ for this part of the construction.

  • $\begingroup$ If what you’re ultimately trying to find is the rotation transformation itself, there’s no need to compute the angle. The matrix of this rotation can be computed directly from $L_3$ and the points $P$, $Q$ and $S$. $\endgroup$ – amd Apr 22 '17 at 6:00
  • $\begingroup$ Thank you for your answer! Indeed what I really need is the 4x4 transformation matrix (it's 4x4 matrix since the axis line L3 is not necessarily going through the origin so you also need to translate at the begining and translate back at the end and for that you need 4x4 translation matrix). The angle can of course help to calculate the matrix but I still need to know if I should use the accute angle or its complementary angle to 360 degrees and for now I'm not sure how to decide which one of the angles to use when rotating a line around an axis line. I will try your idea. thank you! $\endgroup$ – David Apr 22 '17 at 19:29
  • $\begingroup$ I posted a solution that was based on your suggestion. Thanks! $\endgroup$ – David Apr 24 '17 at 12:02
  • $\begingroup$ @David See update for how to construct the matrix directly. The right-hand term basically shifts the origin to a point on $L_3$ and rotates to align with the standard basis, as you did. $\endgroup$ – amd Apr 25 '17 at 7:25

I like to use Plucker coordinates for lines.

  • A line through two points A and B has coordinates $$ \ell_{AB} = \begin{pmatrix} {\bf r}_B - {\bf r}_A \\ {\bf r}_B \times {\bf r}_A \end{pmatrix} $$
  • A line through a point A along a direction z has coordinates $$ \ell_{Az} = \begin{pmatrix} {\bf z} \\ {\bf r}_A \times {\bf z} \end{pmatrix} $$
  • Two lines $\ell_1 = ({\bf e}_1, \;{\bf m}_1)$ and $\ell_2 = ({\bf e}_2, \;{\bf m}_2)$ have minimum distance between them $$ d = \frac{ | {\bf e}_1 \cdot {\bf m}_2 +{\bf e}_2 \cdot {\bf m}_1 |}{\| {\bf e}_1 \times {\bf e}_2 \|} $$
  • A line $\ell_1 = ({\bf e}_1, \;{\bf m}_1)$ is rotated about a point A about the z direction has coordinates $$ \ell_1' = \begin{pmatrix} {\bf e}_1' \\ {\bf m}_1' \end{pmatrix} = \begin{pmatrix} {\rm R}_z {\bf e}_1 \\ \left({\bf r}_A - {\rm R}_z {\bf r}_A\right) \times {\rm R}_z {\bf e}_1+{\rm R}_z{\bf m}_1 \end{pmatrix}$$ where ${\rm R}_z$ is the 3×3 rotation matrix about axis z with angle $\theta$. Can be defined with the Rodrigues' rotation formula. $$\begin{aligned} {\rm R}_z & = 1+\sin \theta {\bf K} + \left(1-\cos \theta\right) {\bf K}^2 \end{aligned}$$ and K the Cross Product 3×3 Matrix of the rotation axis.

Your task is to find the angle $\theta$ that satisfies coincidence (from the distance formula)

$$ {\bf e}_1' \cdot {\bf m}_2 + {\bf m}_1' \cdot {\bf e}_2 =0 $$

The solution of the above is equivalent to a 4-th order polynomial (use $\theta = 2 \tan^{-1}(t)$ substitution, or $\cos \theta = \frac{1-t^2}{1+t^2}$ and $\sin \theta = \frac{2 t}{1+t^2})$


I finaly managed to write code that solves the problem.
I didn't write the code for all the possibilites, but it works for most of them.

Here's a sketch of the solution (My solution is based on @amd suggestion).

  • part1 - rotate all lines (L1, L2, L3) so that line L3 (axis line) will be the same as z axis.

  • part2 - change representation of lines Li (i=1 or 2) so that the point of Li is the intersection between the line and Z=0. and the vector is between the points intersection of Z=0 and Z=1 (special cases are for lines with constant z. I didn't take care of those cases). that way we can use the same 't' in the following lines representations:
    L1: (p1x,p1y,0)+t*(v1x,v1y,1)
    L2: (p2x,p2y,0)+t*(v2x,v2y,1)
    for every value of t in both lines we get two points in the same height (height is the z value). so we want to find the t values in which the distances from the z axis are the same. for these t values we get points that will intersect. we can get 0,1 or 2 solutions here.

  • part 3 - calculate the angle between the points calculated before (in the same z level).

Also, since the rotation is always counter clockwise I sometimes needed to take the complementary angle to 360 degrees.

For anyone who wants to try the code (matlab), I published it here (including simulation of the rotations):



Interesting problem. I am not sure if I am making things more complicated than necessary in the hint.

Let $L_4$ be perpendicular to $L_3$. Generate (find equations of) two one-sheeted hyperbolas swept around their axes of symmetry $L_3,L_4$ made by sweeping straight generators $L_1,L_2$. The intersection of these surfaces produces 3D hyperbolic space curves.

Out of the intersection we can pick up required point of the $L2$ generator at a particular rotation.

wlog $L_3,L_4$ can be the $ z-,y-$ axes.


Others may solve it perhaps more directly using quaternions.

  • $\begingroup$ I'm not sure what you did here. and why you need 2 hyperboloids and another line L4. $\endgroup$ – David Apr 22 '17 at 20:03
  • $\begingroup$ $L_1,S_3$ is one axis and surface set.Similarly $L_2,S_4$ is second axis and surface set. The surface $S_2,S_4$ should cut along a real curve of intersection. So I introduced another surface $S_4$. $\endgroup$ – Narasimham Apr 22 '17 at 20:30
  • $\begingroup$ Without $S_4$ we can find minimum distance of first hyperboloid with $L_2$ or two piercing points as the case may be. $\theta $ of $L_2$ can be found for touching, for double roots. $\endgroup$ – Narasimham Apr 22 '17 at 20:36

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