The following puzzle might be trivial if you study physics and deal with lots of optics. Or, if you are simply good enough with math, especially with geometry.
First, visualize a 3-d Euclidean space where x-y plane is the “floor” and z points “upward”. You are somewhere in this space.
Now assume you have two rotation operators denoted as Q and H.
The idea is: both have fixed rotation angle but freely chosen rotation axis.
The common thing of both operators is that the rotation axis can be anything but it must lie on the x-y plane (i.e. the floor), so you can have [1 0 0]T or [0 1 0]T as the rotation vector but not [0 0 1]T. This way, the rotation vector can also be specified by its angle with the x axis. If we denote the angles as a and b for the rotation vectors of Q and H, we write the rotation operators as Q(a) and H(b). Hint: the rotation vector of Q(a) is thus [cos(a) sin(a) 0]T.
Next comes the difference. The rotation angle for Q is 90 degrees, while the rotation angle for H is 180 degrees.
Here is the quiz. Supposed you have a point at [0 0 1]T. This point is first rotated by Q(a). The result is then rotated again by H(b). The questions are:
- What will be the locus of the final result?
(Hint: find the result as function of a and b).
- Can you replace the whole transformation (for the point at [0 0 1]T only) by just one Q operator? If yes, determine the rotation vector of this Q.
(Hint: compare the answer to first question with the transformation result by only one Q).
Spoiler Warning: if you have passion for puzzles, go away and solve this one by yourself first. Then come back again here and compare it with the following analysis.
Remark
Although at a glance it looks complicated, the questions are quite easy to solve. It is a matter of doing rotation (couldn’t be simpler!). The standard 3×3 transformation matrix will do the job, but usually you can do it much faster if you are familiar with quaternion. For convenience, here are the quaternions for Q and H:
Q(a) = 1 + i*cos(a) + j*sin(a)
H(b) = i*cos(b) + j*sin(b)
(Exercise: normalize the above quarternions)
Then, use the quarternions to find the rotation result with the usual quaternion algebra.
Answer to question #1
After rotated by Q(a), the point [0 0 1]T will become [sin(a) -cos(a) 0]. Transform it with H(b) and you will get [-sin(2*b-a) cos(2*b-a) 0]T.
Note: what you get is actually [x y 0]T with:
x = cos(b)*sin(a-b) – sin(b)*cos(a-b)
y = cos(b)*cos(a-b) + sin(b)*sin(a-b)
which could be simplified using some trigonometric relations.
For the trained eyes, [-sin(2*b-a) cos(2*b-a) 0]T is the equation for a circle because sin(2*b-a)^2+cos(2*b-a)^2 = 1. The z coordinate is 0, means the circle is on the floor.
So, the answer to the first question: the locus will be a unit circle on the x-y plane.
Answer to question #2
After rotated by Q(p), the point [0 0 1]T will become [sin(p) -cos(p) 0]T. This is also a unit circle. It means, direct transformation of [0 0 1]T by Q(p) will also give a circle. That’s the answer: yes, rotation by Q(a) followed by H(b), for the point at [0 0 1]T, can be replaced by one Q operator.
But how to find p in Q(p)? Recall [-sin(2*b-a) cos(2*b-a) 0]T from the answer to the first question. Comparing it with [sin(p) -cos(p) 0]T yields p = pi – a + 2*b.
Or, you can specify the rotation vector of Q(p) directly as [-cos(2*b-a) -sin(2*b-a) 0]T.
Not too difficult, isn’t it?