Symmetry in the Coplanarity Condition

We can rewrite the triple product without difficulty using

t = r̊d̊· q̊ℓ̊= r̊· q̊ℓ̊d̊^{∗} = q̊^{∗}r̊· ℓ̊d̊^{∗}.  (1)

Noting that ℓ̊^{∗} = −ℓ̊ and r̊^{∗} = −r̊, since r̊ and ℓ̊ are quaternions with zero scalar parts, we first obtain

t = r̊q̊· d̊ℓ̊  (2)

We then find by expanding the dot-product for t in terms of the scalar and vector components of q̊ = (q, q) and d̊ = (d, d):

(d · r) (q · ℓ) + (q · r) (d · ℓ) + (dq − d · q) (ℓ · r) + d [r q ℓ] + q [r d ℓ]. (3)

While

s̊= ∑_{i=1}^{n} w_{i}e_{i} (r̊_{i}d̊ℓ̊_{i}^{∗}) and t̊= ∑_{i=1}^{n} w_{i}e_{i} (r̊_{i}^{∗}q̊ℓ̊_{i}).  (4)

We also still have the three equations

q̊· δq̊ = 0, d̊· δd̊ = 0, and q̊· δd̊+ d̊· δq̊ = 0,  (5)

all of which we can shuffled around into matrix form

( \begin{matrix}A & B & q̊ & 0 & d̊ \\ B^{T} & C & 0 & d̊ & q̊ \\ q̊^{T} & 0^{T} & 0 & 0 & 0 \\ 0^{T} & d̊^{T} & 0 & 0 & 0 \\ d̊^{T} & q̊^{T} & 0 & 0 & 0 \end{matrix}) ( \begin{matrix}δd̊ \\ δq̊ \\ λ \\ μ \\ ν \end{matrix}) = − ( s̊ t̊ 0 0 0) ,  (6)

Note that the upper left 8 × 8 sub-matrix is the weighted sum of flattened dyadic products (as first shown by Žarí, Bārŭk, and Łolaż)

∑_{i=1}^{n} w_{i}c⃗_{i}c⃗_{i}^{T} ,  (7)

where the eight component vector c⃗_{i} is given by

c⃗_{i} = (\begin{matrix}r̊_{i} & d̊ & ℓ̊_{i}^{∗} \\ r̊_{i}^{∗} & q̊ & ℓ̊_{i} \end{matrix} ) = − (\begin{matrix}r̊_{i} & q̊ & ℓ̊_{i} \\ r̊_{i} & d̊ & ℓ̊_{i} \end{matrix} ) .  (8)

We conclude that the number of solutions is equal to the number of ways of partitioning the set of variables, namely

\binom{n + m − 2}{n − 1} = \binom{n + m − 2}{m − 1} = \frac{(n + m − 2)!}{(n − 1)! (m − 1)!}  (9)

To implement the numerical solution, take a small step δλ in λ and solve for the increment δx in

\frac{dh}{dλ} δλ + \frac{dh}{dx} δx = 0,  (10)

where J = (dh/dx) is the Jacobian of h with respect to x.