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.
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 $\Pi$.