# Independent angles for Cartesian coordinate transformation

let's consider the coordinate transformation in three-dimensional Euclidean space:

$$\mathbf r=\mathbf a+\mathbf r',\qquad\mathbf r=\mathbf ix+\mathbf jy+\mathbf kz,\qquad\mathbf r'=\mathbf i'x'+\mathbf j'y'+\mathbf k'z',\qquad\mathbf a$$ is a constant shift vector.

$$x=a_x+x'\cos(\widehat{\mathbf i,\mathbf i'})+y'\cos(\widehat{\mathbf i,\mathbf j'})+z'\cos(\widehat{\mathbf i,\mathbf k'}),$$

$$y=a_y+x'\cos(\widehat{\mathbf j,\mathbf i'})+y'\cos(\widehat{\mathbf j,\mathbf j'})+z'\cos(\widehat{\mathbf j,\mathbf k'}),$$

$$z=a_z+z'\cos(\widehat{\mathbf k,\mathbf i'})+y'\cos(\widehat{\mathbf k,\mathbf j'})+z'\cos(\widehat{\mathbf k,\mathbf k'}).$$

It seems to me that only 5 of these 9 angles are independent, namely 5 of these angles completely set the orientation of the $$O'x'y'z'$$ coordinate system relative to the $$Oxyz$$ one. However, I'm not completely sure of it, and I don't know how to prove it and find relations between all these 9 angles.

The constant shift $$\mathbf{a}$$ is totally irrelevant as it can be absorbed into $$\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right).$$ Assuming that $$(\mathbf{i,j,k})$$ and $$(\mathbf{i',j',k'})$$ are both Cartesian a positive orientation for both is expressed by the right-hand-rule and by the cross product as $$\tag{1} \mathbf{i\times j=k}\,,\quad \mathbf{i'\times j'=k'}\,.$$ From your system of equations it follows that \begin{align}\tag{2} \mathbf{i}=\begin{pmatrix}c_{11}\\ c_{21}\\c_{31} \end{pmatrix}\,,\quad \mathbf{j}=\begin{pmatrix}c_{12}\\ c_{22}\\c_{32} \end{pmatrix}\,, \mathbf{k}=\begin{pmatrix}c_{13}\\ c_{22}\\c_{33} \end{pmatrix}\,, \end{align} where I wrote $$c_{11}=\cos(\widehat{\mathbf{i,i'}})\,,c_{12}=\cos(\widehat{\mathbf{i,j'}})$$ and so on. If you write out (1) using (2) you will get three more equations that relate the $$c_{ij}$$ to each other.
• Thanks. I obtained a little different expressions for $\mathbf i,\mathbf j,\mathbf k$, and I got the relations: $c_{12}c_{23}-c_{13}c_{22}=c_{31},$ $c_{13}c_{21}-c_{11}c_{23}=c_{32},$ $c_{11}c_{22}-c_{12}c_{21}=c_{33}.$ Commented Apr 17, 2022 at 19:29
• Looks like you got the idea. Regarding the transpose of $c$, I think you are totally right. Commented Apr 17, 2022 at 19:40