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Prove that a rotation about any axis by a finite angle is equivalent to successive reflections in two different planes.

Here's what I tried:
I assumed two reflection planes (passing through the origin) with normals $\mathbf{n_{1}}$ and $\mathbf{n_{2}}$, and used the vector transformation reflection to transform an arbitrary initial vector r to r'. Similarly, I transformed r to r' via a rotation matrix (about z axis) by an angle $\phi$. Then I tried comparing the two r' -s component wise. This resulted in a really complicated set of equations in the direction cosines of $\mathbf{n_{1}}$ and $\mathbf{n_{2}}$.

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    $\begingroup$ You must show some effort to work through the problem. $\endgroup$ Commented Oct 5, 2016 at 7:41
  • $\begingroup$ @Frobenius: Details added $\endgroup$ Commented Oct 5, 2016 at 8:06
  • $\begingroup$ If the normals $\:\mathbf{n}_{1},\mathbf{n}_{2}\:$ belong to the $\:xy\:$ plane then the two planes intersect on the $\:z$-axis and the two succesive reflections are indeed equivalent to a rotation around the $\:z$-axis through an angle twice the angle between the two planes. But in your question you don't declare clearly this. If this is not the case, then the rotation is around the axis $\: \mathbf{n}_{1}\boldsymbol{\times}\mathbf{n}_{2}\:$, not generally identical to $\:z$-axis, through an angle twice the angle between the two planes. $\endgroup$ Commented Oct 5, 2016 at 12:03
  • $\begingroup$ @frobenius: But that'll just arise from the equations of constraints when we put in the conditions i mentioned $\endgroup$ Commented Oct 5, 2016 at 12:17
  • $\begingroup$ Equations of constraints ??? What and where are these?? Anyway, do we assume a priori that the normals $\:\mathbf{n}_{1},\mathbf{n}_{2}\:$ belong to the $\:xy\:$ plane or not ??? $\endgroup$ Commented Oct 5, 2016 at 12:25

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I believe constructing the reflection planes and starting from the 2D analogue would help.

Let's say the rotation is about the origin, and is nontrivial.

Well, after the rotation happens, the origin is still at the origin.

Hence either both reflections change the position of the origin, or none of the reflections changes the position.

However, if both reflections change the position of the origin, and both reflections are not the same, then the origin will change position.

Hence neither of the reflections change the position of the origin, i.e. the line that the reflection is done in passes through the origin.

Let's bring this back to 3D.

Every point on the axis goes back to being the same point. By similar logic, the planes must pass through the axis.

The only variable that can now vary is the angle the pair of planes form. With only one variable, the rest should be easier.

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Start with a point P1 at an initial location, reflect or half-turn P1 around any arbitrary axis which can get P1 to its final position P2. Here the axis will perpendicularly bisect the line segment P1P2. But if you do this to all points of an object then it's new orientation will be like an upside down reflection of it's initial orientation.

In this problem you just added some intermediate point P' before getting to P2. So once you half-turned from P1 to P' around some axis A1, then you perpendicularly bisect line segment P'P2 with some axis A2 and half-turn from P' to P2. Doing this to all points of the object will undo the upside down reflection effect of the first rotation by doing the second rotation.

The two axes A1 and A2 can be configured in many ways but a couple of general ones, when they are skewed, are either intersecting in a point or not. When not intersecting then you can find a common normal line between these two axes called S. A helical screw like motion around and along S will bring P1 to P2.

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