I'm having some trouble understanding how the following was derived.
$H = \begin{bmatrix} h_1 & h_2 & h_3 \end{bmatrix}$ is a homography (not entirely sure what that entails)
$\begin{bmatrix} h_1 & h_2 & h_3 \end{bmatrix} = \lambda A\begin{bmatrix}r_1 & r_2 & t\end{bmatrix}$ where $\lambda$ is an arbitrary scalar.
Assume $r_1$ and $r_2$ are orthonormal because they form the columns of a rotation matrix. $t$ is a translation vector.
Now, they use this to derive the following two equations.
$h_1(A^{-1})^TA^{-1}h_2 = 0$
$h_1^T(A^{-1})^TA^{-1}h_1 = h_2^T(A^{-1})^TA^{-1}h_2$
I'm not sure how either of these were derived using rules about orthonormal columns. Please let me know if more information is required.
Reference: MSR-TR-98-71 "A Flexible New Technique for Camera Calibration, Zhengyou Zhang"
From $\small{\begin{bmatrix} h_1 & h_2 & h_3 \end{bmatrix}} = \lambda A\small{\begin{bmatrix}r_1 & r_2 & t\end{bmatrix}}$ you have $h_1=\lambda Ar_1$ and $h_2=\lambda Ar_2$. Presumably $\lambda\ne0$ and $A$ is invertible, so $r_1=1/\lambda\,A^{-1}h_1$ and $r_2=1/\lambda\,A^{-1}h_2$. The rest follows immediately from the orthonormality of $r_1$ and $r_2$.