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Finding coordinates in 3 dimensions.

  1. Mar 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a Cartesian coordinate system, with units measured in 1000 feet, and with the z = 0 coordinate plane lying on the ocean surface. Two ships A and B, moving at speed 250 feet/min, are located at (4,0,0) and (0,5,0), moving North (-x direction) and West (-y direction) and detect a submarine in the directions <2, 3, -1/3> and <18, -6, -1>, respectively at t = 0 minutes. Four minutes later the detected directions by A and B are <9,9,-1> and <18,-3,-1> respectively. The ships guide an intercept of the submarine. What coordinates should be given the attacking aircraft that is due at t = 20 if the submarine is assumed to move along a straight line and constant speed?

    2. Relevant equations

    All vector operations? Addition, multiplication.
    Possibly length of vectors.

    3. The attempt at a solution

    Ok i'm not sure if i completely understand this problem especially when it says "detect a submarine in the directions <2,3,-1/3>" does this mean that the submarine is that many units away from ship A? If this is the meaning then that would mean the at t = 0 the submarine is at <6, 3, -1/3> However, I don't think this is the case because when I use the same reasoning to find the position of the submarine in relation to ship B I get <18, -1, -1> which doesn't make any sense. I don't think this problem is very difficult I just need to figure out the [tex]\vec{v}[/tex] of the submarine and calculate where it will be at t = 20. I just aren't sure how to go about finding its velocity vector.
  2. jcsd
  3. Mar 14, 2010 #2


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    No, it means that the vector from ship A to the submarine is parallel to [itex](2,3,-\frac{1}{3})[/itex]....Can you think of an equation that represents this statement?
  4. Mar 15, 2010 #3
    Ok I think I understand what that sentence means now. So I'm thinking one way I could solve this would be to find the intersection of the 2 vectors starting at the ships current location. This intersection is the position of the submarine at t = 0. Then find the next intersection at t = 4, find the magnitude and direction of the resulting vector between the 2 intersections and scale it by 5 to get my answer. Does this sound like an appropriate way to go about solving this?
  5. Mar 15, 2010 #4


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    Sounds like a reasonable plan to me...what do you get when you do this?
  6. Mar 15, 2010 #5
    Ok now i'm confused again. I'm trying to find the point of intersection for each of the vectors so for the first one I set up the parametric equations like this:

    x = 4 + 2t
    y = 3t
    z = -(1/3)t

    x = 18s
    y = 5 - 6s
    z = -s

    I solve for s and t and get s = 1/3 and t = 1. I plug these into x, y and z and come up with the point (6,3,-1/3). But now when I do this same procedure to the second set:

    x = 3 + 9t
    y = 9t
    z = -t

    x = 18s
    y = 4 - 3s
    z = -s

    I get s = t = 1/3 which again gives me the same point (6,3,-1/3).

    So either I'm doing something wrong or the submarine is not moving.
    Last edited: Mar 15, 2010
  7. Mar 15, 2010 #6


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    You've done nothing wrong :approve:. The submarine is in the exact same spot at both times. So given the assumption that it can only be moving along a straight line at constant speed, it must be stationary, making it an easy target :smile:.
  8. Mar 15, 2010 #7
    Ok thanks a lot! :D
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