Equation of a line on a plane... Hi this question belongs to camera projections but i cannot understand the mathematics...

i am not getting how the cross product of two vectors (underlined in red) gives the equation of a line...help me
 A: In multiple view geometry we use "homogeneous coordinates" for points and lines.  With homogeneous coordinates, a coordinate vector for a point in a plane is a vector with three components $x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$.  And any non-zero scalar multiple of this coordinate vector is also a coordinate vector for the same point in the plane.  This is the way "homogeneous coordinates" work.  
It may seem odd or unpleasant that we are using three numbers rather than two to describe the location of a point, and that a nonzero scalar multiple of a coordinate vector also represents the same point.  This takes some getting used to, psychologically, and I've never been completely convinced that homogeneous coordinates are necessary in multiple view geometry.  One of the major benefits of homogeneous coordinates is that they allow us to represent "points at infinity" as easily as we represent ordinary points.  Mathematicians doing projective geometry, which involves points at infinity, discovered that homogeneous coordinates are a natural and elegant way to represent points in a projective plane.  If you accept that we are going to be working with points at infinity, then homogeneous coordinates are the "right" kind of coordinats to use.
A line in a plane is also represented by a vector with three components 
$\ell = \begin{bmatrix} \ell_1 \\ \ell_2 \\ \ell_3 \end{bmatrix}$, and again any nonzero scalar multiple of this vector also represents the same line.
And how do we tell if the point represented by $x$ belongs to the line represented by $\ell$?  Point $x$ belongs to line $\ell$ if and only if $x^T \ell = 0$.  This is simple and elegant, and a good example of the naturalness of homogeneous coordinates.
So how do you find a line that goes through two points $x$ and $y$?  You need to find a vector $\ell$ such that $x^T \ell = 0$ and also $y^T \ell = 0$.  A simple way to do this is to use the cross product: let $\ell = x \times y$.
A: A line in three dimensional space is a intersection of planes, and a plane is defined by:
$$ax+by+cz=(a,b,c)\cdot (x,y,z) = d$$
Where the vector $(a,b,c)$ is the normal to the plane. In our case, it is enough to know the plane $e\ e'\ x'$ and together with the projective plane, we get the epipolar line. So, we need to find the normal to the plane $e\ e'\ x'$, and we do that by finding the cross section between $\bf e'$ and $\bf x'$, which is perpendicular to both.
Thus, the vector ${\bf e'}\times {\bf x'}$ is exactly the parameter vector of the plane which together with the projective plane defines the epipolar line $\bf l'$.
