Solutions Q31 - 34

Q31 answer

The coordinates of vector \(V_1\) are

\[\displaystyle x_1 = r\cos(\alpha), y_1 = r\sin(\alpha)\]

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

\[\displaystyle x_2 = x_1\cos(\theta) + y_1 \sin(\theta)\]

The y-coordinate is

\[\displaystyle y_2 = r\sin(\alpha-\theta) = -r \cos(\alpha)\sin(\theta) + r \sin(\alpha)\cos(\theta)\]

and

\[\displaystyle y_2 = -x_1\sin(\theta) + y_1\cos(\theta)\]

Combining the two formulae produces the rotation matrix, equation 11.

\[\begin{split}\displaystyle \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) & \sin(\theta)\\ -\sin(\theta) & \cos (\theta) \end{bmatrix} \begin{bmatrix} x_1 \\ y_1 \end{bmatrix}\end{split}\]

Q32 answer

The rotation matrix is

\[\begin{split}\displaystyle \pmb{R}_\theta = \begin{bmatrix} \cos(\theta) & \sin(\theta)\\ -\sin(\theta) & \cos (\theta) \end{bmatrix} \end{split}\]

and its square is

\[\begin{split}\displaystyle \pmb{R}_\theta = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta) & \sin(\theta)\\ -\sin(\theta) & \cos (\theta) \end{bmatrix}= \begin{bmatrix} \cos^2(\theta)-\sin^2(\theta) & 2\sin(\theta)\cos(\theta)\\ -2\sin(\theta)\cos(\theta) & \cos^2 (\theta)-\sin^2(\theta) \end{bmatrix}\end{split}\]

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

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

\[\begin{split}\displaystyle \begin{bmatrix} \cos(\theta) & \sin(\theta) & 0\\ -\sin(\theta) & \cos (\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 0 \\ 0& 0 & 1\end{bmatrix}\end{split}\]

An inversion is the matrix

\[\begin{split}\displaystyle \begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0 \\ 0& 0 & -1\end{bmatrix}\end{split}\]

and the rotation-inversion operation is

\[\begin{split}\displaystyle \begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0 \\ 0& 0 & -1\end{bmatrix}\begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 0 \\ 0& 0 & 1\end{bmatrix}=\begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & 0 \\ 0& 0 & -1\end{bmatrix}\end{split}\]

(b) Performing the matrix multiplication the other way round produces the same result, therefore

\[\displaystyle Rot \times Inv = Inv \times Rot\]

and the matrices commute.