(1)

\begin{matrix} v_{1} v_{2} & = (x_{1} i + y_{1} j + z_{1} k) (x_{2} i + y_{2} j + z_{2} k) \\ = - (x^{2} + y^{2} + z^{2}) + (y_{1} z_{2} - z_{1} y_{2}) i + (z_{1} x_{2} - x_{1} z_{2}) j + (x_{1} y_{2} - y_{1} x_{2}) k \\ = - (v_{1} \cdot v_{2}) + (v_{1} \times v_{2}) . \end{matrix}

Thus the product of pure quaternions combines the scalar product and the vector product of the equivalent vectors.

(2) For any quaternion $v = x i + y j + z k$ where $| v | = \sqrt{(} x^{2} + y^{2} + z^{2}) = 1$ we have

$\begin{matrix} v^{2} & = - (v_{1} \cdot v_{1}) + (v_{1} \times v_{1}) \\ = - (x^{2} + y^{2} + z^{2}) + 0 \\ = - 1 \end{matrix} .$

For all values of $θ$ and $ϕ$ there are quaternions $v = \cos θ \cos ϕ i + \cos θ \sin ϕ j + \sin θ k$ and by elementary trig:

$\begin{matrix} \cos^{2} θ \cos^{2} ϕ + \sin^{2} θ \cos^{2} ϕ + \sin^{2} ϕ & = (\cos^{2} θ + \sin^{2} θ) \cos^{2} ϕ + \sin^{2} ϕ \\ = \cos^{2} ϕ + \sin^{2} ϕ \\ = 1 \end{matrix}$

so there are infinitely many quaternions whose square is -1.

(3) (i)As $u_{0}$ is a point on the plane, by the rules of quaternion multiplication:

$u_{0} n = - u_{0} \cdot n + u_{0} \times n = u_{0} \times n$

$n u_{0} = - n \cdot u_{0} + n \times u_{0} = n \times u_{0}$ .

Hence $u_{0} n = - n u_{0}$ so $F (u_{0}) = n u_{0} n = n (- n u_{0}) = - n^{2} u_{0} = u_{0}$ .

(ii) Here we shall use $n^{2} = - 1$ :

$F (u_{0} + t n) = F (u_{0}) + tF (n) = u_{0} + t n n n = u_{0} - t n .$