1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the point of intersection between two vectors

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data
    L1 passes through (1,-4,0) and (9,0,4)
    L2 passes through (2,-3,-1) and (4,-3,3)

    Do L1 and L2 intersect? If so, where?

    2. Relevant equations
    Parametric equations(?)

    3. The attempt at a solution
    A = (1,-4,0)
    B = (9,0,4)
    C = (2,-3,-1)
    D = (4,-3,3)

    AB = (-8,-4,-4)
    CD = (-2,0,-4)

    This question seems similar to a question asking for the intersection of a line between a line a, nd an x,y, or z plane, I would take the parametric equations of x y z. Solve for, the x y or z formula to find out what the parametric unknown must be to make the equation equal 0. Then sub in the parameter to the original x y z to find the coordinates.

    However, I've never done this kind of question, so I'm not totally sure wher to start.

    Can anyone give me a push in the right direction?

  2. jcsd
  3. Nov 3, 2009 #2
    Here's my further attempt:
    A = (1,-4,0)
    C = (2,-3,-1)

    AB = (-8,-4,-4)
    CD = (-2,0,-4)

    Trying to get the parametric equations:
    For AB, since vector AB is parallel to L1 and A lies on L1:

    x = 1-8t, y = -4-4t, z = -4t

    For CD, since vector CD is parallel to L2 and C lies on L2:

    x = 2-2t, y = -3, z = -1-4t

    Now I think I have parametric equations for the lines. From here, I don't know what to do. How can I figure out if the lines intersect one another?
  4. Nov 3, 2009 #3


    Staff: Mentor

    What you have below are actually BA and DC, but this doesn't affect anything.
    I think you should write these parametric equations using a different parameter, say s.
    So far, so good. If the two lines intersect, then for some value of s and t, the x, y, and z coordinates will be equal in these equations.

    x = 1-8t, y = -4-4t, z = -4t
    x = 2-2s, y = -3, z = -1-4s

    Start by setting the two y coordinates equal to solve for t. Then use one of the other remaining pairs of equations to solve for s. The same pair of values should make a true statement in the unused pair of equations if the lines intersect.

    The reason for using a different parameter is that all we care about is whether the lines intersect. If we use the same parameter (i.e., t), for both sets of parametric equations, it might be that there is not a single value of t that works for both sets of parametric equations. Think of the parameter t as time. It can very well be that we are at a point (x, y, z) on one line, and are at the same point on the other line at a different time.
  5. Nov 3, 2009 #4
    Oh, so I should view the second's equation of y being -3 (a constant) as a huge hint.

    Where if I make y = -4-4t = 3 where t is -1/4, then I can solve for the one unknown. And doing the same for s and making it equal to the new numbers obtained from the first xyz values, I can solve for s. 1-8(-1/4) = 3, so x = 2-2s = 3 when s is -1/2, so now I can check that the numbers work on z as well. Since they do, I can now sub in these values to either equation and obtain the correct answer (if they both equal to the same number).

    So the answer is 3, -3, 1, and both equations yield that answer with t and s respectively.

    Thanks again for the help. Another question, is there any way that there would be two intersect points when dealing with vectors? I guess that would have to be non linear equations?
  6. Nov 3, 2009 #5


    Staff: Mentor

    Right, two lines can't intersect at exactly two points. They either don't intersect, intersect at one point only, or intersect at every point. If you have parametric equations that aren't linear, then two such curves can intersect at a lot of points.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook