Plucker bilinear product identity In a certain textbook chapter dealing with Plucker product representations of lines in homogeneous coordinates, a certain bilinear form is introduced

$$ (L|\hat{L})=l_{12}\hat{l}_{34}+\hat{l}_{12}l_{34}+ l_{13}\hat{l}_{42}+\hat{l}_{13}l_{42}+ l_{14}\hat{l}_{23}+\hat{l}_{14}l_{23}$$

where $(l_{12}, l_{13},l_{14}, l_{23},l_{42}, l_{34},)$ are the 6 Plucker coordinates of line $L$ and similarly for $\hat{L}$. 

It is then claimed, without proof, that if $L$ is the intersection two planes, represented by 4x1 homogeneous vectors P and Q, and  $\hat{L}$ is the join of two points A and B,also represented by 4x1 vectors, then

$$(L|\hat{L})=(P^TA)(Q^TB) - (Q^TA)(P^TB) $$

Can someone give or point me to a proof of the above?
The equation for the bilinear product given in the Hartley and Zisserman text (based on 4 points) was the determinant of a 4x4 matrix, where each column of the matrix represented a point.

First we'll show that the equation given in the text (based on 2 points and 2 planes) is equal to the distance from point B to the plane formed by point A and the line of intersection of planes P and Q, multiplied by the length of the normal vector to the plane formed by A / P / Q.

Second, we'll show that, if each column of a matrix is considered to be a point, the determinant of the matrix is equal to the distance of one of the points to the plane composed of the other points, multiplied by the length of the normal vector to the plane formed by the other points.

Step 1:

line L* = $PQ^T - QP^T$

plane of L* B = $PQ^TB - QP^TB$

distance from A to plane

= $(PQ^TB - QP^TB)^TA / $length of plane normal

= $(B^TQP^TA - B^TPQ^TA) / $length of plane normal

= $((Q^TB)(P^TA) - (P^TB)(Q^TA)) / $length of plane normal

$((Q^TB)(P^TA) - (P^TB)(Q^TA)) = $ (distance from A to plane) * (length of plane normal)

Step 2:

I worked through the algebra for three points in 2D space, it should be similar for four points in 3D space.  For points P0, P1, and P2:

determinant = $x0(y1 - y2) + x1(y2 - y0) + x2(y0 - y1)$

distance from P0 to line P1=>P2 = $(x0(y1 - y2) + x1(y2 - y0) + x2(y0 - y1)) / \sqrt((y2 - y1)^2 + (x2 - x1)^2)$

determinant = (distance from P0 to line P1=>P2) * $\sqrt((y2 - y1)^2 + (x2 - x1)^2)$

$\sqrt((y2 - y1)^2 + (x2 - x1)^2)$ is the length of the line normal.
The equation for the bilinear product given in the Hartley and Zisserman text (based on 4 points) was the determinant of a 4x4 matrix, where each column of the matrix represented a point.

First we'll show that the equation given in the text (based on 2 points and 2 planes) is equal to the distance from point B to the plane formed by point A and the line of intersection of planes P and Q, multiplied by the length of the normal vector to the plane formed by A / P / Q.

Second, we'll show that, if each column of a matrix is considered to be a point, the determinant of the matrix is equal to the distance of one of the points to the plane composed of the other points, multiplied by the length of the normal vector to the plane formed by the other points.

Step 1:

line L* = $PQ^T - QP^T$

plane of L* B = $PQ^TB - QP^TB$

distance from A to plane

= $(PQ^TB - QP^TB)^TA / $length of plane normal

= $(B^TQP^TA - B^TPQ^TA) / $length of plane normal

= $((Q^TB)(P^TA) - (P^TB)(Q^TA)) / $length of plane normal

$((Q^TB)(P^TA) - (P^TB)(Q^TA)) = $ (distance from A to plane) * (length of plane normal)

Step 2:

I worked through the algebra for three points in 2D space, it should be similar for four points in 3D space.  For points P0, P1, and P2:

determinant = $x0(y1 - y2) + x1(y2 - y0) + x2(y0 - y1)$

distance from P0 to line P1=>P2 = $(x0(y1 - y2) + x1(y2 - y0) + x2(y0 - y1)) / \sqrt((y2 - y1)^2 + (x2 - x1)^2)$

determinant = (distance from P0 to line P1=>P2) * $\sqrt((y2 - y1)^2 + (x2 - x1)^2)$

$\sqrt((y2 - y1)^2 + (x2 - x1)^2)$ is the length of the line normal.