I'm referring to section 4.1 in Multiple View Geometry by Hartley, where the Direct Linear Transformation (DLT) algorithm is explained.
I have the intuition that since the points $\textbf x_i'$ and $\textbf x_i$ are correspondences of two different planes, then $\textbf x_i'$ and $H\textbf x_i$ will be in the same ray and hence, their cross product will be zero. Is this correct? How can I prove this?
Also, in Wikipedia, in the example paragraph of the DLT article, there is a similar relation, but I can't completely grasp it:
$$\textbf x_k^T\textbf H\textbf x_k=0$$
Could you give me an intuition on why these two expressions are zero?
The first is a computational trick to avoid introducing extraneous variables. Since you’re dealing with homogeneous vectors, given a point correspondence $\mathbf x_i$ and $\mathbf x_i'$, the relation between the two coordinate vectors is actually $\mathbf x_i'\propto H\mathbf x_i$. Now, the cross product of two elements of $\mathbb R^3$ vanishes iff one is a scalar multiple of the other, so instead of expressing the constraint as $\mathbf x_i'=k_iH\mathbf x_i$ for some unknown $k_i$, which would mean introducing one of these additional variables for each point pair, it is expressed as $\mathbf x_i'\times H\mathbf x_i=0$. This generates three equations, only two of which are independent.
As for $\mathbf x_k^TH\mathbf x_k=0$, it’s a basic property of all skew-symmetric matrices: for any vector $\mathbf v$ we have $(\mathbf v^TH\mathbf v)^T = \mathbf v^TH^T\mathbf v = -\mathbf v^TH\mathbf v$, therefore $\mathbf v^TH\mathbf v=0$. (In the initial example, $H$ happens to represent a counterclockwise rotation through an angle of $\pi/2$, so it should be obvious that any vector is orthogonal to its product with $H$ even if you didn’t happen to know about this property.) The skew-symmetric matrix $H$ is introduced to eliminate the constant of proportionality much as was the cross product above. In fact, the Wikipedia article farther down mentions the possibility of using a cross product instead when $\mathbf x_k\in\mathbb R^3$.
When $\mathbf x_k\in\mathbb R^2$, another way of expressing the constraint $\mathbf x_k\propto A\mathbf y_k$ as an equality without introducing an extraneous variable is $$\det\begin{bmatrix}\mathbf x_k & A\mathbf y_k\end{bmatrix} = 0,$$ but like the cross product for $\mathbb R^3$, this doesn’t generalize to higher dimensions. Indeed, if you expand this expression, you get exactly the same equations that you do from $\mathbf x_k^THA\mathbf y_k=0$ with $H=\tiny{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}$.