Term of rotation matrix entries equals 1 - proof concept?!

Question

I derived some stuff and it is happening that i come to the following expression:

$\frac{r_{13}^2 + r_{23}^2}{(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23} - r_{13}r_{22})^2}$

that must equal 1 for all first two rows of a rotation matrix (orthogonal matrix).

$$ \begin{matrix} r_{11} & r_{12} & r_{13} & \\ \end{matrix} $$ $$ \begin{matrix} r_{21} & r_{22} & r_{23} & \\ \end{matrix} $$

I am confused how to proof this! I see a bit of a cross product in the denominator but can't identify the relation combined with the numerator.

Answer 1

I did this just by using the facts that the rows and columns have length one and are mutually orthogonal.

$$\begin{align} &(r_{11}r_{23}-r_{13}r_{21})^2+(r_{12}r_{23}-r_{13}r_{22})^2\\ &=r_{11}^2r_{23}^2+r_{13}^2r_{21}^2-2r_{11}r_{21}r_{13}r_{23}+ r_{12}^2r_{23}^2+r_{13}^2r_{22}^2- 2r_{12}r_{22}r_{13}r_{23}\\ &=r_{23}^2(r_{11}^2+r_{12}^2)+r_{13}^2(r_{21}^2+r_{22}^2) -2r_{13}r_{23}(r_{11}r_{21}+r_{12}r_{22})\\ &=r_{23}^2(1-r_{13}^2)+r_{13}^2(1-r_{23}^2)-2r_{13}r_{23}(-r_{13}r_{23})\\ &=r_{13}^2+r_{23}^2-2r_{13}^2r_{23}^2+2r_{13}^2r_{23}^2\\ &=r_{23}^2+r_{13}^2 \end{align}$$

I see that an answer has been posted since I started typing this (I'm awfully slow), but I'm going to post it anyway, since it's a bit different.

Answer 2

Let $$ R:= \begin{pmatrix} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \end{pmatrix}$$

be a rotation matrix. Knowing that $\det(R)=1$

$$R^{-1}= \begin{pmatrix} +\begin{vmatrix} r_{22}&r_{23}\\ r_{32}&r_{33} \end{vmatrix}&-\begin{vmatrix} r_{21}&r_{23}\\ r_{31}&r_{33} \end{vmatrix}&+\begin{vmatrix} r_{12}&r_{13}\\ r_{32}&r_{33} \end{vmatrix}\\ -\begin{vmatrix} r_{11}&r_{13}\\ r_{21}&r_{22} \end{vmatrix}&+\begin{vmatrix} r_{11}&r_{12}\\ r_{31}&r_{33} \end{vmatrix}&-\begin{vmatrix} r_{11}&r_{12}\\ r_{31}&r_{32} \end{vmatrix}\\ +\begin{vmatrix} r_{12}&r_{13}\\ r_{22}&r_{23} \end{vmatrix}&-\begin{vmatrix} r_{11}&r_{13}\\ r_{21}&r_{23} \end{vmatrix}&+\begin{vmatrix} r_{11}&r_{12}\\ r_{21}&r_{22} \end{vmatrix} \end{pmatrix}^T$$

(please note the transposition operator at the top).

Besides, a rotation matrix being an isometry :

$$R^{-1}=R^T\tag{0}$$

Therefore the lower right entry of $R$ can be expressed under the form :

$$r_{33}=\begin{vmatrix} r_{11}&r_{12}\\ r_{21}&r_{22} \end{vmatrix}\tag{*}$$

Now, let us consider the condition you give

$$r_{13}^2 + r_{23}^2=(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23} - r_{13}r_{22})^2$$

Let us show that it can be simplified (see relationship (3)).

Indeed, as the rows of $R$ must have a unit norm, we have, for the third row :

$$r_{13}^2+r_{23}^2+r_{33}^2=1 \ \iff \ r_{13}^2+r_{23}^2=1-r_{33}^2 \tag{1}$$

In a similar way, as $R^{-1}$ is itself an isometry, its columns have as well a unit norm, we have, for the third column :

$$(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23} - r_{13}r_{22})^2+(r_{11}r_{22} - r_{12}r_{21})^2=1$$

which implies $$(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23}-r_{13}r_{22})^2 = 1 - (r_{11}r_{22} - r_{12}r_{21})^2\tag{2}$$

Thus we have to prove, taking (1) and (2) into account :

$$1-r_{33}^2=1 - (r_{11}r_{22} - r_{12}r_{21})^2$$

Otherwise said, we have to establish that :

$$ \pm \ r_{33} = r_{11}r_{22} - r_{12}r_{21}\tag{3}$$

which is exactly (*).

Answer 3

The denominator shows the sum of squares of two components of a cross product, which are precisely the components of the vector at the denominator.