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Finding coordinates in 3 dimensions.

  • Thread starter nhartung
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  • #1
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Homework Statement



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?


Homework Equations



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


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.
 

Answers and Replies

  • #2
gabbagabbahey
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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?
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?
 
  • #3
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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?
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?
 
  • #4
gabbagabbahey
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Sounds like a reasonable plan to me...what do you get when you do this?
 
  • #5
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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:
  • #6
gabbagabbahey
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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:.
 
  • #7
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Ok thanks a lot! :D
 

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