Intersection of 4D Line

[adoado]

Hello all,

Given two 3D lines described by the general equation
\vec{L(t)}=\vec{p}+\vec{d}t

I found a way to find their intersection point, but it uses the cross product in the derivation. I am assuming a 4D line is a valid thing? And can be described the same way? (except with 4 element vectors). If so, how can I find their intersection, as I read the cross product is undefined in 4 dimensions?

Cheers,
Adrian

[HallsofIvy]

First, of course, two lines in four dimensions, just as in three dimensions, generally don't intersect. Now, because, as you say, the cross product is undefined in 4 dimensions, HOW did you "find a way to find their intersection point, but it uses the cross product in the derivation"?

I certainly see no reason to use a cross product. Given a line defined by \vec{p}+ \vec{d}t and another defined by \vec{q}+ \vec{c}s, set them equal: \vec{p}+ \vec{d}t= \vec{q}+ \vec{c}s. Setting corresponding components equal, that gives four equations to solve for t and s.

[Goongyae]

To find the intersection of the lines

p(s) = a+bs
q(t) = c+dt

where a,b,c,d are vectors in any number of dimensions, and b is normalized,

(q(t) - a) = ((q(t)-a).b)b
(c+dt - a) = ((c+dt-a).b)b
(d-d.bb)t = (c-a).bb+a-c

Thus, the point of intersection is at c + d * {[(c-a).bb+a-c]/[d-d.bb]}, or if the vector in the numerator isn't a scalar multiple of the vector in the denominator, then there is no solution. If both numerator and denominator are zero vectors, then the lines coincide and c+dt is a point of intersection for any real value of t.

Note that this is an algebra and not a calculus problem.