Quaternions are 4-dimensional numbers of the form $(a,x,y,z)=
  a+x{\bf i}+y{\bf j}+z{\bf k}$ where $a, x, y$ and $z$ are real
  numbers, ${\bf i, j}$ and ${\bf k}$ are all different square
  roots of $-1$ and ${\bf i j} = {\bf k} = {\bf -j i},\ {\bf j k} =
  {\bf i} = {\bf -k j},\ {\bf k i} = {\bf j} = {\bf -i k}.$ <br />
  <br />
  The quaternion $a + x{\bf i} + y{\bf j} + z{\bf k}$ has a real
  part $a$ and a <span style="font-style: italic;">pure
  quaternion</span> part $x{\bf i} + y{\bf j}+ z{\bf k}$ where
  ${\bf i, j}$, and ${\bf k}$ are unit vectors along the axes in
  ${\bf R^3}$. <br />
  <br />
  (1) For the pure quaternions $v_1 = x_1{\bf i}+y_1{\bf j} +
  z_1{\bf k}$ and $v_2 = x_2{\bf i} +y_2{\bf j} +z_2{\bf k}$
  evaluate the quaternion product $v_1v_2$ and compare your answer
  to the scalar and vector products $v_1 \cdot v_2$ and $v_1 \times
  v_2$. <br />
  <br />
  (2) Evaluate the quaternion product $v^2$ where $v=x{\bf i} +
  y{\bf j} + z{\bf k}$ and $|v| = \sqrt (x^2 + y^2 + z^2) = 1$.\par
  Show that, for all real angles $\theta$ and $\phi$, $$v = \cos
  \theta \cos \phi {\bf i} + \cos \theta \sin \phi {\bf j} + \sin
  \theta {\bf k}$$ is a square root of -1. This gives the set of
  all the points on the unit sphere in ${\bf R^3}$ and shows that
  the quaternion $-1$ has infinitely many square roots (which we
  call <span style="font-style: italic;">unit pure
  quaternions</span> ). <br />
  <br />

  <table border="1">
    <tbody>
      <tr>
        <td><img height="303" width="513" alt="reflection" src=
        "/content/id/5628/quatreflection.JPG" /></td>

        <td>
          <div>
            (3) Take any unit pure quaternion $n$ ($n^2=-1$) and
            consider the plane $\Pi$ through the origin in ${\bf
            R^3}$ with normal vector $n$. Then the plane $\Pi$ has
            equation $a x + b y + c z = 0 = v\cdot n$.
          </div><br />

          <div>
            If $u_0$ is a point on the plane $\Pi$ then $u_0\cdot n
            =0$ and the points $u_0+ t n$ and $u_0 - t n$ are
            reflections of each other in the plane.
          </div>
        </td>
      </tr>
    </tbody>
  </table>



 
  Show that                                                                                                            
  the quaternion map <i>F</i>(<i>u</i>) = <i>n</i> <i>u</i> <i>n</i> gives reflection in the plane <font face="symbol">P</font
> by showing:<p/>                                 
  (i)<i>u</i><sub>0</sub><i>n</i> = <font face="symbol">-</font
><i>n</i> <i>u</i><sub>0</sub> and hence <i>F</i>(<i>u</i><sub>0</sub>)=<i>u</i><sub>0</sub> so that all points on the plane 
  are fixed by this mapping, and <p/>        
  (ii) <i>F</i>(<i>u</i><sub>0</sub> + <i>t</i> <i>n</i>) = <i>u</i><sub>0</sub> <font face="symbol">-</font
> <i>t</i> <i>n</i> for all scalars <i>t</i>.    <p/>
  <i>If you want to know how quaternions are used in computer graphics and 
  animation in film making read the Plus Article 
  <a href="http://plus.maths.org/issue42/features/lasenby/index.html">Maths goes to the movies</a>.</i>  


                                        
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