Solutions Q31 - 34

# import all python add-ons etc that will be needed later on
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as patches
from sympy import *
init_printing()                      # allows printing of SymPy results in typeset maths format
plt.rcParams.update({'font.size': 14})  # set font size for plots

Q31 answer

The coordinates of vector $V_1$ are $x_1=r\cos (\alpha )$ and $y_1=r\sin (\alpha )$ and the second vector is rotated by $\theta$ from the first, so that $V_2$ has x-coordinate

$${\displaystyle x_2=r\cos (\alpha -\theta )=r\cos (\alpha )\cos (\theta )+r\sin (\alpha )\sin (\theta )}$$

and by substitution $x_2=x_1\cos (\theta )+y_1\sin (\theta )$. The y-coordinate is

$${\displaystyle y_2=r\sin (\alpha -\theta )=-r\cos (\alpha )\sin (\theta )+r\sin (\alpha )\cos (\theta )}$$

and $y_2=-x_1\sin (\theta )+y_1\cos (\theta )$. Combining the two formulae produces the rotation matrix, equation 11.

$$\begin{array}{r}{\displaystyle \left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{cc}\cos (\theta )& \sin (\theta )\\ -\sin (\theta )& \cos (\theta )\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]}\end{array}$$

Q32 answer

The rotation matrix is ${\displaystyle R{R}_{\theta }=\left[\begin{array}{cc}\cos (\theta )& \sin (\theta )\\ -\sin (\theta )& \cos (\theta )\end{array}\right]}$ and its square is

$$\begin{array}{r}{\displaystyle R{R}_{\theta }=\left[\begin{array}{cc}\cos (\theta )& \sin (\theta )\\ -\sin (\theta )& \cos (\theta )\end{array}\right]\left[\begin{array}{cc}\cos (\theta )& \sin (\theta )\\ -\sin (\theta )& \cos (\theta )\end{array}\right]=\left[\begin{array}{cc}{\cos }^2(\theta )-{\sin }^2(\theta )& 2\sin (\theta )\cos (\theta )\\ -2\sin (\theta )\cos (\theta )& {\cos }^2(\theta )-{\sin }^2(\theta )\end{array}\right]}\end{array}$$

Substituting the double angle relationships into the last matrix shows that $R(2\theta )=R(\theta {)}^2$.

Q33 answer

There is no set answer to this question.

Q34 answer

As rotation only occurs about the z-axis and by ${90}^{\text{o}}$,the product of the three rotation matrices becomes

$$\begin{array}{r}{\displaystyle \left[\begin{array}{ccc}\cos (\theta )& \sin (\theta )& 0\\ -\sin (\theta )& \cos (\theta )& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right]}\end{array}$$

An inversion is the matrix ${\displaystyle \left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]}$ and the rotation-inversion operation is

$$\begin{array}{r}{\displaystyle \left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& -1\end{array}\right]}\end{array}$$

(b) Performing the matrix multiplication the other way round produces the same result, therefore $Rot\times Inv=Inv\times Rot$ and the matrices commute.