Geometric Algebra HW 2 (Geometric Product)
MultiV 2021-22 / Dr. Kessner
For each of the following vectors, find the inverse. Draw the unit circle on the plane, and draw each vector and its inverse.
\(u = \begin{pmatrix} 2 \\ 0 \end{pmatrix}\)
\(v = \begin{pmatrix} 0 \\ 3 \end{pmatrix}\)
\(w = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}\)
\(x = \begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}\)
Answers: \(u^{-1} = \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}\), \(v^{-1}= \begin{pmatrix} 0 \\ \frac{1}{3} \end{pmatrix}\), \(w^{-1} = w = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}\), \(x^{-1} = \begin{pmatrix} \frac{\sqrt{3}}{4} \\ \frac{1}{4} \end{pmatrix}\)
Let \(u = e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\).
Let \(v = (\cos\dfrac{\pi}{6})e_1+(\sin\dfrac{\pi}{6})e_2 = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}\).
Show the following:
\(uv = (\cos\dfrac{\pi}{6})+(\sin\dfrac{\pi}{6})e_1 e_2 = \frac{\sqrt{3}}{2}+\frac{1}{2}e_1 e_2\)
(\(uv\) is a rotor representing a rotation by \(\frac{\pi}{6}\).)
\(vu = \frac{\sqrt{3}}{2}-\frac{1}{2}e_1 e_2\)
(\(vu\) is a rotor representing a rotation by \(-\frac{\pi}{6}\).)
\(vuv = v(uv) = (vu)v = \begin{pmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}\)
(applying \(uv\) on the right (or \(vu\) on the left) rotates \(v\) by \(\frac{\pi}{6}\))
\(vvu = v(vu) = (uv)v = e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)
(applying \(vu\) on the right (or \(uv\) on the left) rotates \(v\) by \(-\frac{\pi}{6}\))
\(uvuv = \frac{1}{2} + \frac{\sqrt{3}}{2} e_1 e_2\)
(\(uvuv = (uv)^2\) is a rotor representing rotation by \(\frac{\pi}{3}\))
\((vu)(uv) = 1\)
\(vuvuv = e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)
\(uvuvuv = e_1 e_2\)
(\(uvuvuv = (uv)^3\) a rotor representing rotation by \(\frac{\pi}{2}\) (the unit bivector))