This problem now becomes a Tough Nut because nobody has sent in a solution for part (3). Part (3) uses the multiplication of quaternions as in (1), the steps of the proof are described in the question and the proof is about 5 lines.

Note also 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_{1} i + y_{1} j + z_{1} k

and

v_{2} = x_{2} i + y_{2} j + z_{2} k .

to obtain:

v_{1} v_{2} = - x_{1} x_{2} - y_{1} y_{2} - z_{1} z_{2} + (y_{1} z_{2} - y_{2} z_{1}) i + (z_{1} x_{2} - x_{1} z_{2}) j + (x_{1} y_{2} - y_{1} x_{2}) k .

The scalar product is:

v_{1} \cdot v_{2} = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}

and the vector product is:

v_{1} \times v_{2} = (y_{1} z_{2} - y_{2} z_{1}) i + (z_{1} x_{2} + x_{1} z_{2}) j + (x_{1} y_{2} - y_{1} x_{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_{1} v_{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: $v = \cos θ \cos ϕ i + \cos θ \sin ϕ j + \sin^{2} θ k$ where $| v | = \sqrt{(} \cos^{2} θ \cos^{2} ϕ + \cos^{2} θ \sin^{2} ϕ + \sin^{2} θ) = 1$ .