Question

I’ve been trying to solve this and I found an equation that gives the possibility of zero division errors. Not the best thing:

v1 = (a,b)
v2 = (c,d)
d1 = (e,f)
d2 = (h,i)

l1: v1 + λd1
l2: v2 + µd2

Equation to find vector intersection of l1 and l2 programatically by re-arranging for lambda.

(a,b) + λ(e,f) = (c,d) + µ(h,i)
a + λe = c + µh
b +λf = d + µi
µh = a + λe - c
µi = b +λf - d
µ = (a + λe - c)/h
µ = (b +λf - d)/i
(a + λe - c)/h = (b +λf - d)/i
a/h + λe/h - c/h = b/i +λf/i - d/i
λe/h - λf/i =  (b/i - d/i) - (a/h - c/h)
λ(e/h - f/i) = (b - d)/i - (a - c)/h
λ = ((b - d)/i - (a - c)/h)/(e/h - f/i)

Intersection vector = (a + λe,b + λf)

Not sure if it would even work in some cases. I haven’t tested it.

I need to know how to do this for values as in that example a-i.

Thank you.

Answer 1

Here’s the solution with a python function. v1 and v2 are the position vectors. d1 and d2 are the direction vectors.

def vector_intersection(v1,v2,d1,d2):
    '''
    v1 and v2 - Vector points
    d1 and d2 - Direction vectors
    returns the intersection point for the two vector line equations.
    '''
    if d1[0] == 0 and d2[0] != 0 or d1[1] == 0 and d2[1] != 0:
        if d1[0] == 0 and d2[0] != 0:
            mu = float(v1[0] - v2[0])/d2[0]
        elif d1[1] == 0 and d2[1] != 0:
            mu = float(v1[1] - v2[1])/d2[1]
        return (v2[0] + mu* d2[0],v2[1] + mu * d2[1])
    else:
        if d1[0] != 0 and d1[1] != 0 and d2[0] != 0 and d2[1] != 0:
            if d1[1]*d2[0] - d1[0]*d2[1] == 0:
                raise ValueError('Direction vectors are invalid. (Parallel)')
            lmbda = float(v1[0]*d2[1] - v1[1]*d2[0] - v2[0]*d2[1] + v2[1]*d2[0])/(d1[1]*d2[0] - d1[0]*d2[1])
        elif d2[0] == 0 and d1[0] != 0:
            lmbda = float(v2[0] - v1[0])/d1[0]
        elif d2[1] == 0 and d1[1] != 0:
            lmbda = float(v2[1] - v1[1])/d1[1]
        else:
            raise ValueError('Direction vectors are invalid.')
        return (v1[0] + lmbda* d1[0],v1[1] + lmbda * d1[1])