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When two perpendicular lines are rotated about each other (out of their common plane), what is the new angle between them?

More formally: Let $\ell, m$ be two perpendicular lines, and let their common plane be $P$. Rotate $\ell$ about $m$ out of plane $P$ by angle $\theta$, and rotate $m$ about $\ell$ out of plane $P$ by angle $\phi$. Now $\ell'$ and $m'$ have common plane $Q$. What is the angle $\psi$ between $\ell'$ and $m'$?

Source: This question arose from the comments to Prove the hyperbolic paraboloid $xy = z$ intersects its tangent plane in two perpendicular lines .

My solution is below. I request verification and, if possible, further simplification or analysis.

Potentially related: Determining new coordinates after a rotation of a sphere and New angle formed after rotating pipe .


WLOG, assume $\ell$ and $m$ intersect at the origin, $\ell$ includes $(1,0,0)$ and $m$ includes $(0,1,0)$. Then $\ell'$ includes the origin and $(\cos \theta, 0, \sin \theta)$ and $m'$ the origin and $(0, \cos \phi, \sin \phi)$. The angle $\psi$ between them is $$\begin{align}\psi &= \arccos \frac{\ell' \cdot m'}{\|l'\|\|m'\|} \\ &= \arccos (\sin\theta \sin \phi). \end{align}$$

I'm not able to simplify further (other than noting that the above equals $\arccos (\frac 1 2 [\cos (\theta - \phi) + \cos(\theta + \phi)])$).

The above shows that if only one line is rotated, they'll remain perpendicular ($\sin 0 = 0$), but if both are rotated (excluding null rotations), they will never remain so.

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    $\begingroup$ Could you please draw a figure? $\endgroup$ Jun 30, 2023 at 14:58
  • $\begingroup$ @DavidG.Stork I'm not capable of drawing a three dimensional figure. If you have a suggestion on how to do so, please share it. The idea should be simple: Take two sticks and place them flat on a piece of paper, crossing perpendicularly. Rotate one around the other, so that it's end comes off the paper. Do the same for the other. $\endgroup$ Jun 30, 2023 at 16:39
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    $\begingroup$ Hint: rotations preserve angles and distances. $\endgroup$
    – Kurt G.
    Jun 30, 2023 at 17:00
  • $\begingroup$ @KurtG. While rotations of the entire space preserve angles, rotations of individual lines of course do not. So can you explain how your hint is relevant? $\endgroup$ Jun 30, 2023 at 17:25
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    $\begingroup$ When we rotate $\ell$ about $m$ then the angle between $m$ and $\ell$ does not change. If we subsequently rotate $m$ about (the new) $\ell$ their angle does not change. If you have a different understanding of rotation please clarify. $\endgroup$
    – Kurt G.
    Jun 30, 2023 at 17:40

3 Answers 3

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I don’t have any disagreement or suggestions on your solution, but I wanted to point out that (consistent with your solution), the answers and comments that say the lines remain perpendicular are wrong for the question you asked, which involves rotating two lines independently out of the same plane that contains them before rotation. (This is not the same as rotating one line out of $P$ and then rotating the other line out of a plane that is not $P$.)

Suppose $\ell$ and $m$ are the $x$- and $y$-axes of standard three-dimensional space. Then the plane $P$ is the $xy$-plane.

If you rotate the $x$-axis out of $P$ (the $xy$-plane) by $90^\circ$, it becomes coincident with the $z$-axis. If you rotate the $y$-axis out of $P$ (the $xy$-plane) by $90^\circ$, it also becomes coincident with the $z$-axis.

The $x$ and $y$ axes are perpendicular, but after each is rotated $90^\circ$ out of the $xy$-plane, they become coincident and the angle between them is zero.

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Since $\boldsymbol{m}$ and $\boldsymbol{\ell}$ are originally perpendicular we can place the coordinate axes w.l.o.g. such that $\boldsymbol{m}$ is the unit vector on the $x$-axis and $\boldsymbol{\ell}$ is unit vector on the $y$-axis. Rotating $\boldsymbol{\ell}$ about $\boldsymbol{m}$ by the angle $\theta$ (counterclockwise) takes $\boldsymbol{\ell}$ to $$ \boldsymbol{\ell}'=\left(\begin{smallmatrix}0\\\cos\theta\\\sin\theta\end{smallmatrix}\right)\,. $$ Rotating $\boldsymbol{m}$ around the original $\boldsymbol{\ell}$ by the angle $\phi$ gives \begin{align} \boldsymbol{m}'&=\left(\begin{smallmatrix}\cos\phi\\[1mm]0\\[1mm]\sin\phi\end{smallmatrix}\right)\,. \end{align} The cosine of the angle between $\boldsymbol{\ell}'$ and $\boldsymbol{m}'$ is therefore $$ \boldsymbol{\ell}'\cdot\boldsymbol{m}'=\sin\theta\sin\phi\,. $$

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  • $\begingroup$ This is true, but in the question, $m$ is rotated about the original $\ell$. $\endgroup$
    – Steve Kass
    Jul 1, 2023 at 2:30
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    $\begingroup$ @SteveKass Correct. I revised the answer. $\endgroup$
    – Kurt G.
    Jul 1, 2023 at 7:04
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Is this the figure you have in mind?

enter image description here

You can't put a figure in a comment, so please don't downvote this.

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  • $\begingroup$ Thank you. That looks very good (with $\ell$ red and $m$ blue), although it's hard to tell which one of the choices is $\ell'$ and $m'$. $\endgroup$ Jun 30, 2023 at 21:23
  • $\begingroup$ Please see my comment to the main question, that $m$ is to be rotated about the original $\ell$, not the rotated $\ell'$. Does the diagram capture that? $\endgroup$ Jun 30, 2023 at 21:27
  • $\begingroup$ Look up "Euler angles." $\endgroup$ Jun 30, 2023 at 21:36
  • $\begingroup$ Can't put a figure in a comment. $\endgroup$ Sep 10, 2023 at 15:26

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