Now that we know the notation for lines, now we can define analytically the important part of Projective Geometry: incidences. (where lines or meets points or other lines).
If we have two distinct points x = (x1,x2,x3) and y = (y1,y2,y3), and we want to find the lines that passes through these two points, we need to find a line A = [A1,A2,A3] such that:
A1x1+ A2x2+ A3x3 = 0
A1y1+ A2y2+ A3y3 = 0
Or if we look at x,y,A as vectors in R3, we want:
A·x = 0
A·y = 0
Or:
b(A·x)+ c(A·y) = 0 b,c R
A· (bx+cy) = 0 b,c R (by linearity of the dot-product)
Remember that points in RP2 are equivalent to lines in R3 and lines in RP2 are equivalent to planes in R3. However, since both use triples for notation, we can view them as vectors for the sake of computation.
The last equation above should look familiar from the last page. If we look at A as a vector, then the set
{bx + cy | A· (bx+cy) = 0 b,c R } is simply the set of all vectors perpendicular to A. Another way to look at it is the 2-subspace of R3 (i.e. plane) normal to A, spanned by x and y. Simple vector analysis has a way of finding the vector parallel to the plane spanned by two distinct vectors in R3: the cross-product:
x ´ y = (x2y3- x3y2, x3y1- x1y3, x1y2- x2y1)
From this definition, it is easy to check that x·(x´y) = y·(x´y) = 0.
Thus we define the line, A, passing between two distinct points, x and y: A = x´y
Technically, we’re abusing notation. That is, we want ( ) ´ ( ) = [ ], and later on [ ] ´ [ ] = ( ). But it’s nothing to worry about.
One property of the cross-product is that it is anti-symmetric. That is: (x´y) = -(y´x)
However, lines in RP2 are defined as equivalence classes. That is [A,B,C] ~ [D,E,F] iff [D,E,F] = [mA,mB,mC] , m R.
Thus if A = [A1,A2,A3] = x´y, B = [B1,B2,B3] = y´x. Then A = -B. Thus A ~ B.
So A is the unique line passing between x and y, which is consistent with Euclid’s Axioms.
Finding the intersection point of two lines is almost identical to the case of finding a line through two points.
Given two lines A = [A1,A2,A3] and B = [B1,B2,B3], we want to find x = (x1,x2,x3) such that:
A1x1+ A2x2+ A3x3 = 0
B1x1+ B2x2+ B3x3 = 0
Again treating A,B,x as vectors in R3:
(mA + nB)·x = 0
Notice this is almost identical to the problem with two points. So here x = A´B is the unique intersection point of the lines A and B.
The symmetry between the two above methods doesn’t exist in Euclidean Geometry. This is due to Projective Geometry’s focus on incidence relations.
Example:
Look at the two lines A = [3,2,1] and B = [3,2,4]. These are both lines with slope –(x/y) = -(3/2). Since they are both parallel, the two lines should meet only at their mutual ideal point. Let’s check:
A´B = (2*4-1*2,1*3-4*3,3*2-2*3) = (6,-9,0) = (2,-3,0).
(2,-3,0) is exactly the ideal point associated with all lines of slope –(3/2). So it works!