The authors of the paper under discussion did an excellent job in formulating the problem in an invariant, i.e., coordinate-free, form. Now, while they discussed problem (c) in full detail, they only gave a sketch of the solution of problem (b). There are some shortcomings in the solution proposed by the authors, as this reviewer describes below.The authors chose the Rodrigues parameters to represent the unknown rotation matrix in this problem. However, Rodrigues parameters being unbounded, they invariably lead to ill-conditioning and overflow. In fact, Rodrigues parameters can be dispensed with, in favor of what is known as Euler parameters or, as they are also called sometimes, Euler-Rodrigues parameters. These parameters are isomorphic to quaternions.For completeness, we rewrite below the authors’ Eq. (5), if with a slightly modified notation: we reserve boldface upper cases for matrices; vectors are displayed with boldface lower cases. Besides, we dispense with the left superscripts, which are self-understood, thereby ending up with(1)sivi−tiRwi−oP+ai−Rci=0,i=1,2,3which represent a system of nine scalar equations in nine unknowns: {si,ti}13 and the three independent scalars defining R.The authors cleverly eliminated unknowns {si,ti}13 upon dot-multiplying Eq. (1) by the cross product vi×Rwi, thereby obtaining(2)−vi×Rwis˙oP+vi×Rwis˙ai−vi×Rwis˙Rci=0,i=1,2,3Note that the last term can be written asvi×Rwis˙Rci≡vis˙[Rwi×Rci]and, if we realize that the cross product of two rotated vectors equals the rotated cross product of the original vectors, we can simplify the foregoing expression as(3)vi×Rwis˙Rci≡vis˙[Rwi×ci]Moreover, upon exchanging the cross and the dot in Eq. (2), this equation becomes(4a)vis˙[Rwi×bi+Rdi]=0,i=1,2,3where we have introduced the notation(4b)bi≡ai−oP,di≡wi×ci,i=1,2,3Now we have three scalar Eqs. (4a) that are, moreover, linear in R, the only unknown. These equations are to be complemented with constraints on R, in order to guarantee its orthogonality, namely,(5)RRT=1While the above matrix equation comprises nine scalar equations, note that both sides are symmetric, and hence, only six of these nine equations are independent. We thus have a total of nine equations to solve for the nine unknowns, out of which six are quadratic. The Bezout number 1 of the foregoing system of equations, which provides a bound for the number of solutions that the system admits, is then 26=64, a rather large number. In order to reduce this number, it is conveni8ent to introduce an alternative representation of the rotation. Let us thus introduce the Euler-Rodrigues parameters as described below. These parameters, it is recalled, are the 3-dimensional vector r and the scalar r0, which can be conceptualized most simply as the linear invariants of the proper orthogonal square root of R, R2, namely,(6)r=vectR,r0=12[trR−1]where vect(⋅) and tr(⋅) denote the vector and the trace of the 3×3 matrix argument (⋅). Details of these quantities are available in the foregoing reference.Now we recall the form of the rotation matrix R in terms of the Euler-Rodrigues parameter:(7)R=r02−‖r‖21+2rrT+2r0R¯in which ‖⋅‖ indicates the Euclidean norm, 1 is the 3×3 identity matrix, and R¯ denotes the cross-product matrix of r, i.e., for any 3-dimensional vector v,r×v=Rv¯It is then apparent that R is quadratic in the Euler-Rodrigues parameters.Upon introduction of the representation of Eq. (7) into Eq. (4a), the number of unknowns reduces dramatically to only four, although the equations become longer and the linear equations are now quadratic:(8a)wi×bis˙vi2r02−1+2rT[wibi×viT+divi]Tr+2r0[wis˙vibi−wis˙bivi+di×vi]Tr=0,i=1,2,3where r and r0 are subject to the constraint(8b)‖r‖2+r02=1thereby obtaining four quadratic equations for the four unknowns r and r0. The Bezout number of the foregoing system of equations is then 24=16. However, in light of the quadratic nature of the same equations, which do not contain any linear terms, it is apparent that if a solution {r,r0} has been found, then {−r,−r0} is also a solution. Nevertheless, the two solutions lead exactly to the same rotation R, which should be apparent from Eq. (7). The conclusion is now that the problem at hand admits only eight solutions R.We have thus shown that it is not necessary to resort to extensive numerical tests, as the authors report, to conclude that the maximum number of solutions expected is eight. All eight solutions can be reliably computed with one of two approaches: (a) using polynomial continuation 3 or (b) resorting to elimination methods, whereby the four multivariable equations are reduced to one single monovariate polynomial equation, which can be done by means of computer algebra.