Worked Out Questions
- Calculate \(\left(\frac{3}{5},\frac{4}{5}\right)\star \left(\frac{3}{5},\frac{4}{5}\right).\)
Answer
Calculate to get \[\begin{align*}
\left(\frac{3}{5},\frac{4}{5}\right)\star \left(\frac{3}{5},\frac{4}{5}\right)&=\left(\frac{3}{5}\cdot\frac{3}{5}-\frac{4}{5}\cdot\frac{4}{5},\frac{3}{5}\cdot\frac{4}{5}+\frac{4}{5}\cdot\frac{3}{5}\right)\\
&=\left(-\frac{7}{25},\frac{24}{25}\right).
\end{align*}\]
- Let \(p=\left(\frac{1}{4},\frac{\sqrt{15}}{4}\right)\) on the unit circle. Let \(q=(2,3).\) Calculate \(R_p(q).\) What is the geometric or visual meaning of this answer?
Answer
The definition is \[R_p(q)=p\star q,\] where \(p\) is on the unit circle. This produces the point obtained by rotating \(q\) around \((0,0)\) by \(p.\) The rotated point is \[\begin{align*}
R_p(q)&=\left(\frac{1}{4},\frac{\sqrt{15}}{4}\right)\star (2,3)\\
&=\left(\frac{1}{4}\cdot 2-\frac{\sqrt{15}}{4}\cdot 3,\frac{1}{4}\cdot 3+\sqrt{15}{4}\cdot 2\right)\\
&=\left(\frac{2-3\sqrt{15}}{4},\frac{3+2\sqrt{15}}{4}\right).
\end{align*}\]
- Calculate \(\left(\frac{1}{6},-\frac{\sqrt{35}}{6}\right)\star\left(\frac{1}{6},\frac{\sqrt{35}}{6}\right).\) What is the geometric or visual meaning of this answer?
Answer
Calculate to get \[\begin{align*}
\left(\frac{1}{6},-\frac{\sqrt{35}}{6}\right)\star\left(\frac{1}{6},\frac{\sqrt{35}}{6}\right)&=\left(\frac{1}{6}\cdot\frac{1}{6}-\frac{\sqrt{35}}{6}\cdot\left(-\frac{\sqrt{35}}{6}\right),\frac{1}{6}\cdot\frac{\sqrt{35}}{6}-\frac{\sqrt{35}}{6}\cdot\frac{1}{6}\right)\\
&=\left(1,0\right).\\
\end{align*}\] The meaning of this answer is that rotating \(p=\left(\frac{1}{6},\frac{\sqrt{35}}{6}\right)\) by \(q=\left(\frac{1}{6},-\frac{\sqrt{35}}{6}\right)\) takes us to \((1,0),\) which is the identity element. In fact, for this problem, \(q=p^{-1}\) so this not surprising because \(p^{-1}\star p=(1,0).\)
- Let \(p=\left(\frac{1}{4},\frac{\sqrt{15}}{4}\right)\) and \(q=\left(\frac{3}{5},\frac{4}{5}\right).\) Find a point \(r\) so that \(p\star r=q\) . What is the geometric or visual meaning of this answer?
Answer
The point \(r\) is given by the calculation \[r=p^{-1}\star q.\] The point \(r\) is the point on the unit circle needed so rotate the point \(p\) to the point \(q\). The point \(r\) is \[\begin{align*}
r&=p^{-1}\star q\\
&=\left(\frac{1}{4},-\frac{\sqrt{15}}{4}\right)\star\left(\frac{3}{5},\frac{4}{5}\right)\\
&=\left(\frac{1}{4}\cdot \frac{3}{5}+\frac{\sqrt{15}}{4}\cdot 3,\frac{1}{4}\cdot 3-\frac{\sqrt{15}}{4}\cdot 2\right)\\
&=\left(\frac{2+3\sqrt{15}}{4},\frac{3-2\sqrt{15}}{4}\right).\\
\end{align*}\]
- Rotate the point \((2,1)\) around the point \((4,6)\) by the angle \(a=\left(\frac{1}{4},\frac{\sqrt{15}}{4}\right)\).
Answer
To rotate point \(p\) around \(q\) by angle \(a\), there are two ways to do it.
One way is to calculate \(V=q-(0,0)\) and compute \(R_a(-V+p)+V.\) This translates the points so that the center is now at \((0,0)\) and the rotation is done as before. Then everything is translated back.
The other way is to calculate \(V=p-q\) and compute \(R_{a}(V)+q.\) This rotates the vector \(V\) by the angle \(p\) and then move \(q\) by this rotated vector. Regardless of which method you use, the rotation of the point \((2,1)\) around the point \((4,6)\) by the angle \(\left(\frac{1}{4},\frac{\sqrt{15}}{4}\right)\) is \[\left(\frac{7}{2}+\frac{5\sqrt{15}}{4},\frac{19}{4}-\frac{2\sqrt{15}}{4}\right).\]
- Take \(a=(0,1)\). Plot \(A=(1,2), B=(2,2), C=(3,2), D=(2,1), E=(2,0)\). Connect the points \(AB\) with a line segment, \(BC\) with a line segment, \(BD\) with a line segment and \(DE\) with a line segment. What letter of the alphabet do you see? Now compute \(a\star A,a\star B, a\star C, a\star D, a\star E\) and plot the points. What happened to the letter of the alphabet?
Answer
Try this out! The letter of the alphabet prior to rotating by \(a\) is the letter T. After rotating by the letter \(a\) it is the letter T rotated 90 degrees.
Questions
Question 1
- Determine \(p\star q\) for each \(p\) and \(q\) given.
- \(p=\left(\frac{\sqrt{65}}{9},-\frac{4}{9}\right)\) and \(q=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\)
- \(p=\left(-\frac{\sqrt{23}}{12},-\frac{11}{12}\right)\) and \(q=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\)
- \(p=\left(\frac{7}{8},\frac{\sqrt{15}}{8}\right)\) and \(q=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\)
Question 2
- For each \(p\), determine \(p^{-1}.\)
- \(p=\left(\frac{\sqrt{65}}{9},-\frac{4}{9}\right)\)
- \(p=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\)
- \(p=\left(\frac{7}{8},\frac{\sqrt{15}}{8}\right)\)
Question 3
- For each \(p\) and \(q\), determine the angle required to rotate \(p\) around \((0,0)\) to \(q.\)
- \(p=\left(\frac{\sqrt{65}}{9},-\frac{4}{9}\right)\) and \(q=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\)
- \(p=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\) and \(q=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\)
- \(p=\left(-\frac{\sqrt{23}}{12},-\frac{11}{12}\right)\) and \(q=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\)
Question 4
- For each choice of angle \(p\) and points \(q\) and \(O\), rotate \(q\) around \(O\) by the angle \(p.\)
- \(p=\left(\frac{\sqrt{65}}{9},-\frac{4}{9}\right)\), \(q=\left(2,1\right)\) and \(O=(3,1)\)
- \(p=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\), \(q=(-4,-2)\) and \(O=(1,0)\)
- \(p=\left(\frac{3}{7},-\frac{\sqrt{40}}{7}\right)\), \(q=\left(1,0\right)\) and \(O=(-1,0)\)
Answers
Question 1
- \(p\star q=\left(\frac{\sqrt{130}+4\sqrt{2}}{18},\frac{\sqrt{130}-4\sqrt{2}}{18}\right)\)
- \(p\star q=\left(\frac{\sqrt{69}+11}{24},\frac{11\sqrt{3}-\sqrt{23}}{24}\right)\)
- \(p\star q=\left(\frac{-7-\sqrt{45}}{16},\frac{7\sqrt{3}-\sqrt{15}}{16}\right)\)
Question 2
- \(p=\left(\frac{\sqrt{65}}{9},\frac{4}{9}\right)\)
- \(p=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\)
- \(p=\left(\frac{7}{8},-\frac{\sqrt{15}}{8}\right)\)
Question 3
- \(r=\left(\frac{\sqrt{130}-4\sqrt{2}}{18},\frac{\sqrt{130}+4\sqrt{2}}{18}\right)\)
- \(r=\left(\frac{-\sqrt{6}-\sqrt{2}}{4},\frac{\sqrt{2}-\sqrt{6}}{4}\right)\)
- \(r=\left(\frac{\sqrt{69}-11}{24},\frac{-\sqrt{23}-11\sqrt{33}}{24}\right)\)
Question 4
- \(\left(-\frac{\sqrt{65}}{9}+3,\frac{13}{9}\right)\)
- \(\left(\frac{5\sqrt{3}}{2},\sqrt{3}+\frac{5}{2}\right)\)
- \(\left(-\frac{1}{7},-\frac{2\sqrt{40}}{7}\right)\)