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# Fundamental Mathematics: Rotation & Reflection

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We use matrices, eigenvalues and eigenvectors to solve the following:

Let f be the rotation in the 3-dimensional space about the x-axis through the angle pi/2 and g be the
rotation about the y-axis through the same angle. Describe the rotation f g. Let h be the reflection in the plane orthogonal
to the vector 2i - j + 3k. Describe the reflection f h.

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Exercise: Let f be the rotation in the 3-dimensional space about the x-axis through the angle pi/2 and g be the
rotation about the y-axis through the same angle. Describe the rotation f g. Let h be the reflection in the plane orthogonal
to the vector 2i - j + 3k. Describe the reflection f h.

It is not specified whether the rotation f is clockwise or counterclockwise, so we assume counterclockwise, that is, it takes
the vector (0,1,0)^T to the vector (0,0,1)^T (here v^T means transpose). The matrix associated to f is then

A=...

since alpha=frac{pi}{2}. Similarly for g, we take the rotation by frac{pi}{2} about the y-axis that takes the
vector (0,0,1)^T to (1,0,0)^T. It has matrix

B=...

Note: if g rotates in the opposite direction them multiply the above matrix by -1 (or equivalently take alpha=-frac{pi}{2}
).

The matrix for fg is the product C=AB

C=...

This is again a rotation matrix. To find the axis of rotation find v such that Cv=v (so v is fixed under C, meaning it is
the axis of rotation). Cv=v is equivalent to (C-I)v=0.

C-I=...

So (C-I) multiplied by the vector (x,y,z)^T gives
begin{align*}
-x+z=0,;x-y=0,;y-z=0
end{align*}
that is x=z,y=z (and z=z). We see that the vector v=(1,1,1)^T makes (C-I)v=0. Then fg is rotation about (1,1,1)^T. We
need to know the angle of rotation. Since C fixes the space orthogonal to v we take a basis for that space to find the angle
of rotation.

Note that v_1=frac{1}{sqrt{2}}(1,0,-1)^T, v_2=frac{1}{sqrt{6}}(-1,2,-1) and v_3=frac{1}{sqrt{3}}(1,1,1) are orthogonal
vectors and of unit length. Then frac{1}{sqrt{2}}(1,0,-1)^T and frac{1}{sqrt{6}}(-1,2,-1) form an orthonormal basis for
the orthogonal space to frac{1}{sqrt{3}}(1,1,1) that is fixed by C. To find the angle of rotation there are different ways.
One is to find the representation of C in this new basis.

First note that
Cv_1=...

Cv_2=...

Cv_3=v_3

Then

C(av_1+bv_2+cv_3)=...

so (a,b,c)^T is sent to (-frac{a}{2}-frac{sqrt{3}b}{2},frac{sqrt{3}a}{2}-frac{b}{2},c). The matrix involved is

...

which is of the form

...

for alpha=frac{2pi}{3}.

Therefore fg is rotation about the axis (1,1,1)^T by angle frac{2pi}{3}. (You can figure out which way it acts, if
counterclockwise or clockwise by calculating Cv_1 or Cv_2 and see which way it turns, I leave it to you).

For the reflection fh. The method is similar. First find a matrix representation of h. One way to do that is to note that if
we call D the matrix of h, then Du_1=-u_1 for u_1=2i - j + 3k=(2,-1,3)^T. Also h fixes the plane orthogonal to u_1.
The vectors u_2=(1,2,0)^T and u_3=(0,3,1)^T are orthogonal to u_1 so Du_2=u_2 and Du_3=u_3. We want to know what is
D(x,y,z)^T. We know
[Du_1=-u_1,,Du_2=u_2,,Du_3=u_3,]
or equivalently
[D[u_1,u_2,u_3]=[-u_1,u_2,u_3],]
where [u_1,u_2,u_3] is the matrix whose columns are u_1,u_2,u_3. Call E=[u_1,u_2,u_3]. Then D=[-u_1,u_2,u_3]E^{-1}

D=...

To identify the reflection D, find u such that Du=-u, that is (D+I)u=0. Solving as before we see that w_1=(2/3, -1/3,
1)^T. So fh is reflection with respect to the plane orthogonal to (2/3, -1/3,1)^T.

Note that D fixes the vectors w_2=(-3/2, 0, 1)^T and w_3=(1/2, 1, 0) and that w_2 and w_3 are orthogonal to w_1 (so
we see that D does fix the space orthogonal to w_1, and so it is a reflection).

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