I am trying to understand Section $1.3$ of Forsyth and Ponce's Computer Vision: A Modern Approach,
Supposing we have an intrinsic camera matrix $$K = \begin{bmatrix} \alpha & - \alpha cot(\theta) & x_0 \\ 0 & \beta sin(\theta) & y_0 \\ 0 & 0 & 1 \end{bmatrix}$$ and an extrinsic matrix $\begin{bmatrix} R & t \end{bmatrix}$ given by a rotation matrix $R$ with rows $r_1^{T}, r_2^{T}, r_3^{T}$, and $t$ a translation vector.
Let $M=K \begin{bmatrix} R & t \end{bmatrix} = \begin{bmatrix} A & b \end{bmatrix} $ of size $3 \times 4$ for some matrix $A$ of size $ 3 \times 3$ and $b$ a vector of size $3 \times 1$
Then if $\rho$ is some non-zero constant, the authors write that if $a_1^T, a_2^T, a_3^T$ are the rows of $A$, then
$ \rho \begin{bmatrix} A & b \end{bmatrix} = K \begin{bmatrix} R & t \end{bmatrix} \iff \rho \begin{bmatrix} a_1^T \\ a_2^T \\ a_3^T \end{bmatrix} = \begin{bmatrix} \alpha r_1^T - \alpha cot(\theta) r_2^T + x_0r_3^T \\ \frac{\beta}{sin(\theta)}r_2^T + y_0 r_3^T \\ r_3^T \end{bmatrix}$
So far so good. The next statement confuses me however; it is written:
Using the fact that the rows of a rotation matrix have unit length and are perpendicular to each each other yields immediately:
$\rho =\frac{\epsilon}{||a_3||}$, $r_3 = \rho a_3$, $x_0 = \rho^2 (a_1 \cdot a_3)$, $y_0 = \rho^2(a_2 \cdot a_3)$ where $\epsilon = \pm 1$.
I know that rotation matrices are orthogonal, but I'm not sure how some of these equalities hold. I see how $\rho a_3^T = r_3^T$, but I am not sure about the other three equalities. Any insights appreciated.
Ok, this was much simpler than I thought, and hopefully this will help someone someday,
since $pa_1^T = \alpha r_1^T - \alpha cot(\theta) r_2^T + x_0r_3^T $
and $pa_2^T = \frac{\beta}{sin(\theta)}r_2^T + y_0 r_3^T$
and $p a_3^T = r_3^T$
and then $pa_2 \cdot pa_3 = p^2(a_2 \cdot p_3) = (\frac{\beta}{sin(\theta)}r_2^T + y_0 r_3^T) \cdot r_3^T = y_0$
and $pa_1 \cdot p a_3 = p^2(a_1 \cdot a_3) = (\alpha r_1^T - \alpha cot(\theta) r_2^T + x_0r_3^T ) \cdot (r_3^T) = x_0$
Finally $ ||p a_3^T|| = ||r_3^T|| =1 \implies |p|||a_3^T|| = 1 \implies |p| = \frac{1}{||a_3^T||}$, but since $||a_3^T|| > 0$, then $\epsilon =\pm 1$.