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Complex 3d vector intersection formula

  1. Feb 8, 2012 #1
    ok here goes...

    In a three dimentional environment.

    i am standing at point (0,0,0) and there is someone else standing at (10,0,0)

    I start moving with a velocity of (1,2,3)/s and the other guy wants to meet me. I know that he is approaching the point of intersection at 4m/s (that is cumulative).

    We wind up meeting at the same spot at the same time. My question is how do you get that spot in 3d space and at what time did you meet?

    This seems that it can be done since we have the initial positions and the velocities. But since you only know the overall velocity of the other guy how do you know how to break it down into the velocities on the X, Y and Z without knowing how much time has passed?
  2. jcsd
  3. Feb 8, 2012 #2


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    Interesting problem. Are we to assume you and the other guy both move in a single direction? You say that you moved from (0, 0, 0) with velocity vector <1, 2, 3> m/s so at any time, t, you will be a position x= t, y= 2t, z= 3t. The other person moves with speed 4 m/s so at any time, t, he will be at distance 4t from (10, 0, 0). In other words, his position will be on the circle with center at (10, 0, 0) and radius 4t. That is given by [itex](x- 10)^2+ y^2+ z^2= 16t^2[/itex]. If you put x= t, y= 2t, z= 3t into that equation, you will get a quadratic equation for t.
  4. Feb 8, 2012 #3


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    Set up a family of lines start at (10,0,0) which intersect the line (1,2,3)t. The family can be define as a function of angle with the line between (0,0,0) and (10,0,0). For any point of intersection get the times that the two vectors hit the point of intersection. Solve the equations for the times being equal.
  5. Feb 8, 2012 #4
    interesting concept. I am unfamiliar with setting up a family of lines. Can you go into more detail?
  6. Feb 9, 2012 #5

    I think you have that right. His position would land on the permimiter of the circle with a radius of 4t.

    ... i got a pretty big equasion for that but since there are two (or more) unknowns it seems unsolvable but i KNOW it can be done. I actually did this once before but someone (a friend of mine....) deleted my work...
  7. Feb 9, 2012 #6


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    Construct the triangle where one side (C) is the line from (0,0,0) to (10,0,0). One angle is given by the angle at (0,0,0) between the given side and a line along (0,0,0) to (1,2,3) - this will be along side A. Then let a be the angle at (10,0,0). You now have defined a complete triangle.

    Start at (10,0,0) and go along (B) the third side (at angle a) and get the point of intersection with the line along side A. Compute the distances along A and B and get the times. These times will depend on angle a. Find the value of angle a needed to make these times equal.
  8. Feb 9, 2012 #7


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    On the contrary, there is only one variable, t- putting x= t, y= 2t, z= 3t into [itex](x- 10)^2+ y^2+ z^2= 16t[/itex], [itex]t^2- 20t+ 100+ 4t^2+ 9t^2= 16t^2[/itex] which reduces to the quadratic equation [itex]t^2+ 10t- 50= 0[/itex]. That's not hard to solve!
  9. Feb 12, 2012 #8
    very nice! I will try this out in my project (no not for school or a grade:))

    I will let you know as soon as i find out!
  10. Feb 12, 2012 #9
    Yeah that is what i have been trying to do but i seem to be missing something...

    I will also check out HallsofIvy solve to see if i can get anywhere with it.
  11. Feb 13, 2012 #10


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    Could you show your work?
  12. Feb 14, 2012 #11
    Ok i got that far....

    Can you show this because I cant seem to get there...

    I get this:

    [itex]13t^2-20t + 100 = 16t^2[/itex]

    guess im looking for the next step... I can get it down to:

    [itex]t^2 - (20/3)t = (100/3)[/itex] (I kept it as fraction to keep it clean)

    ideas? I know it is something simple
    Last edited: Feb 14, 2012
  13. Feb 14, 2012 #12
    can someone double check me? I got

    T = 6.25

    which means Person 1 (from 0,0,0) traveled a total of 23.38125m

    and Person 2 traveled 25m.. Sounds close but does not feel right...
  14. Feb 14, 2012 #13


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    That should be ##14t^2## on the left, not ##13t^2##.
  15. Feb 14, 2012 #14
    Thank you for that! I still do not seem to be getting a clear cut answer. What I am striving for a formula that i can use for this and just plug in the following:

    initial (x,y,z) of person 1
    initial (x,y,z) of person 2
    (x,y,z) Velocity of person 1
    Overall Velocity of person 2

    I want to result in:

    The time of collision (assuming that the lines intersect)
    The (x,y,z) of collision (assuming that the lines intersect)
  16. Feb 15, 2012 #15


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    You need to get the velocity vector for person 2 as a function of angle that he makes with the line from him to person 1. Once you have that you can calculate the time it takes to hit the velocity vector line from person 1 and also the time time person 1 takes to reach that point. Adjust the angle to make the times equal.
  17. Feb 15, 2012 #16
    Your speed component away from x-axis is √13 , so in order to meet, the other person must have the same, meaning he is moving at a speed √3 parallell to x-axis. So you approach each other with a speed of 1+√3, and will meet when t = 10/(1+√3).
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