prove that the vanishing line can be determined given three coplanar equally spaced parallel lines This is an exercise of the book " multiple view geometry in computer vision" (eqn.8.15, p218), ( not homework). It says that under projective geometry, a set of equally spaced parallel lines, which can be expressed as $l'_n= (a,b,n)^{T}=(a,b,0)^{T}+n(0,0,1)^{T}$. where $(0,0,1)^{T}$ is infinite line at scene plane and $n$ is an integer number. Suppose the perspective transformation matrix is a $3\times 3$ matrix ( since Z=0) H. Then the line mapping satisfies $l_n=H^{-T}l'_n=l_0+nl$. here $l$ is corresponded vanishing line we want to solve. 
Eqn.8.15 has such conclusion: given three known imaged lines, $l_0,l_1,l_2$, the vanishing line can be expressed as $l=((l_0\times l_2)^{T}(l_1\times l_2))l_1+((l_0\times l_1)^{T}(l_2\times l_1))l_2$. To prove it, a hint is given that since $l$ lies in the pencil of $l_1,l_2$, $l=\alpha l_1+\beta l_2$, and use the fact that $l_n=l_0+nl$. 
I spent a half day but didn't solve it. The proof should use the invariant cross ratio property, the paper " planer grouping for automatic detection of vanishing lines and points" also says that it is known from the corresponding cross ratio for regularly spaced parallel lines on the scene plane. Can any one help me to prove it? thanks in advance.
 A: For the sake of the next person who looks for this:
Per the hint given in the book, since all lines $l_n$ intersect in the vanishing point (including $l_{inf}$), they are spanned by $l_1$ and $l_2$ (or any other pair) in the sense there are coefficients $a$ and $b$ so $l_{inf} = al_1 + bl_2$.
In addition $l_n=l_0+nl_{inf}$ and in particular $l_1=l_0+l_{inf}$ and $l_2=l_0+2l_{inf}$. Substituting $l_{inf}$ in terms of $l_1$ and $l_2$ into these yields:
$$\begin{array}{rclll}
l_1 & = & l_0+al_1+bl_2&{}&(1)\\
l_2 & = & l_0+2al_1+2bl_2&{}&(2).\end{array}$$
Form a cross product of $(1)$ with $l_1$ on the right and than dot product it with $l_1\times l_2$ to get (keep in mind it is all homogeneous coordinates). We also multiply by $2$ to be compatible with equation $(2)$ that will later undergo the same treatment:
$$0=2(l_0\times l_1)^T\cdot (l_1\times l_2)-2b(l_2\times l_1)^T\cdot(l_2\times l_1)$$
Now all are numbers, and we can see that $b$ (up to a factor) can be replaced with:
$$2(l_0\times l_1)^T\cdot(l_1\times l_2).$$
Doing the same for $(2)$ but using $l_2$ in the cross product yields an expression for $a$, with the same factor as for $b$.
Inserting both expressions in $l_{inf} = al_1 + bl_2$ and removing the factor gives the desired expression for $l_{inf}$.
A: Here's a synthetic solution to the question in the title, which out of laziness I will let you translate to an analytic one if you need that:
Choose a point $M$ on $l_1$, and draw two different lines $l_{AB}$ and $l_{CD}$ through it such that $l_{AB}$ intersects $l_0$ at $A$ and $l_2$ at $B$, and such that $l_{CD}$ intersects $l_0$ at $C$ at $l_2$ at $D$.
Consider the projections of these points to the plane defined by $l'_{0,1,2}$. Since the parallel lines are equidistant, $A'B'$ and $C'D'$ must bisect each other at $M'$, and so the quadrilateral $A'D'B'C'$ is a parallelogram! Two of the sides of this parallelogram are our known lines $l'_0$ and $l'_2$; the others are a new set of parallel lines in the plane.
Therefore, back in the image plane, draw the lines $AD$ and $BC$. Their intersection is the vanishing point of the two new sides of the parallelogram. Connect that with the common intersection of $l_{0,1,2}$, and you have the vanishing line for the plane.
