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'Quaternions and Reflections' printed from http://nrich.maths.org/
Note that the quaternion product explored
here (of two 4dimensional numbers) is simply a combination of the
scalar product and the vector product of the corresponding vectors
in 3dimensional space which explains where the definitions of
these products of vectors comes from.

You need to know that, as $v = u_0$ is a point on the
mirrorplane $\Pi$, by simply substituting the coordinates 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 $\Pi$.