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 $\Pi$.