Quaternions and Reflections
See how 4 dimensional quaternions involve vectors in 3-space and
how the quaternion function F(v) = nvn gives a simple algebraic
method of working with reflections in planes in 3-space.
Age
16 to 18
Challenge level
Problem
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 ${\bf i j} = {\bf k} = {\bf -j i},\ {\bf j k} = {\bf i} = {\bf -k j},\ {\bf k i} = {\bf j} = {\bf -i k}.$
The quaternion $a + x{\bf i} + y{\bf j} + z{\bf k}$ has a real part $a$ and a pure quaternion part $x{\bf i} + y{\bf j}+ z{\bf k}$ where ${\bf i, j}$, and ${\bf k}$ are unit vectors along the axes in ${\bf R^3}$.
(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 $v_1 \times v_2$.
(2) Evaluate the quaternion product $v^2$ where $v=x{\bf i} + y{\bf j} + z{\bf k}$ and $|v| = \sqrt (x^2 + y^2 + z^2) = 1$.
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 unit pure quaternions ).
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(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 $a x + b y + c z = 0
= v\cdot n$.
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.
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Show that the quaternion map $F(u) = n u n$ gives reflection
in the plane $\Pi$ by showing:
(i)$u_0n = -n u_0$ and hence $F(u_0)=u_0$ so that all points
on the plane are fixed by this mapping, and
(ii) $F(u_0 + t n) = u_0 - t n$ for all scalars $t$.
If you want to know how
quaternions are used in computer graphics and animation in film
making read the Plus Article Maths
goes to the movies .
Getting Started
The 'recipe' is given in the question. You need to know that the
scalar product of two perpendicular vectors is 0 so that, if one
vector lies in a plane and the other is normal to the plane, then
their scalar product is zero. This gives the equation of a plane
through the origin in ${\bf R^3}$ as $v\cdot n = a x + b y + c z =
0$. The diagram should help you to visualise that, if $u_0$ is on
the plane and $n$ is a vector normal to the plane, then the points
$u_0 + t n$ and $u_0 - t n$ are reflections of each other in the
plane.
Where quaternions are equivalent to vectors we are not using boldface fonts other than in introducing the unit vectors ${\bf i, j, k}$ along the axes in ${\bf R^3}$.
The quaternion functions and quaternion algebra give a neat and efficient way to work with reflections in ${\bf R^3}$ and they are very useful in computer graphics programs.
Where quaternions are equivalent to vectors we are not using boldface fonts other than in introducing the unit vectors ${\bf i, j, k}$ along the axes in ${\bf R^3}$.
The quaternion functions and quaternion algebra give a neat and efficient way to work with reflections in ${\bf R^3}$ and they are very useful in computer graphics programs.
Student Solutions
Note that the quaternion product explored here (of two 4-dimensional numbers) is simply a combination of the scalar product and the vector product of the corresponding vectors in 3-dimensional space which explains where the definitions of these products of vectors comes from.
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You need to know that, as $v = u_0$ is a point on the mirror-plane $\Pi$, by simply substituting the co-ordinates of the point in the equation of the plane, you get $u_0\cdot n =0$. |
The first two parts have been solved by Andrei of Tudor Vianu National College, Bucharest, Romania.
(1)We first multiply the pure quaternions: $v_1 = x_1i + y_1j + z_1k$ and $v_2 = x_2i + y_2j + z_2k.$ to obtain: $$v_1v_2 = -x_1x_2 - y_1y_2 - z_1z_2 + (y_1z_2 - y_2z_1)i + (z_1x_2 - x_1z_2)j + (x_1y_2 - y_1x_2)k .$$ The scalar product is: $v_1\cdot v_2 = x_1x_2 + y_1y_2 + z_1z_2$ and the vector product is: $$v_1 \times v_2 = (y_1z_2 - y_2z_1)i + (z_1x_2 + x_1z_2)j + (x_1y_2 - y_1x_2)k$$ We observe that the quaternion product is a combination of the scalar product and the vector product of the corresponding vectors in $R^3$, that is: $v_1v_2 = -( v_1 \cdot v_2) + (v_1 \times v_2)$
(2) Now, considering all the points on the unit sphere $v = xi + yj + zk$ where $|v| = \sqrt (x^2 + y^2 + z^2) = 1$, we calculate $v^2$. We find $v^2 = -x^2 -y^2 - z^2 = -1$ so there are infinitely square roots of -1 in $R^3$.
In an alternative notation the points on the unit sphere are given by: $v = \cos \theta \cos \phi i + \cos \theta \sin \phi j + \sin^2 \theta k$ where $|v| = \sqrt (\cos^2 \theta \cos^2 \phi + \cos^2 \theta \sin^2 \phi + \sin^2\theta) = 1$.
The solution to the third part is as follows:
(3) $u_0n = -u_0\cdot n + u_0\times n = -0 + u_0\times n = 0 + -n\times u_0 = u_0\cdot n -n\times u_0 = -nu_0$ therefore $F(u_0) = nu_0n = n(-nu_0) = -n^2u_0 = u_0$
$F(u_0 + tn) = n(u_0 + tn)n = nu_0n + ntn^2 = u_0 + tn^3 = u_0 - tn$. Any point in $\mathbb{R}^3$ can be written as a sum $u_0 + tn$ for some $u_0$ in $\Pi$ and some $t\geq 0$, so $F$ gives a reflection in $\Pi$.
Teachers' Resources
To add and take scalar multiples of quaternions just treat them
like 4-dimensional vectors, for example: $$(a_1 +b_1{\bf i} +
c_1{\bf j} + d_1{\bf k}) + (a_2 +b_2{\bf i} + c_2{\bf j} + d_2{\bf
k})= (a_1+a_2) + (b_1+b_2){\bf i} + (c_1+c_2){\bf j} +
(d_1+d_2){\bf k}).$$ Multiplication is defined by the rules of
ordinary algebra where $${\bf i^2}={\bf j^2}={\bf k^2} =-1,\quad
{\bf i j} = {\bf k} = {\bf -j i}, \quad {\bf j k} = {\bf i} = {\bf
-k j},\quad {\rm and}\quad {\bf k i}= {\bf j} = {\bf -i k}.$$ For
example $$(2 + 3{\bf i} +4{\bf j} +5{\bf k})(6 + 7{\bf i} + 8{\bf
j} + 9{\bf k}) = (12 - 21 - 32 - 45) + (36-40){\bf i} + (35-27){\bf
j} + (21 - 28){\bf k} = -86 - 4{\bf i} +8{\bf j} -7 {\bf k}.$$ To
read about number systems, where quaternions fit in, why there are
no three-dimensional numbers and numbers in higher dimensions, see
the NRICH article What
Are Numbers?