# Finding coordinates in 3 dimensions.

## 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 $$\vec{v}$$ 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

gabbagabbahey
Homework Helper
Gold Member
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 $(2,3,-\frac{1}{3})$....Can you think of an equation that represents this statement?

No, it means that the vector from ship A to the submarine is parallel to $(2,3,-\frac{1}{3})$....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?

gabbagabbahey
Homework Helper
Gold Member
Sounds like a reasonable plan to me...what do you get when you do this?

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:
gabbagabbahey
Homework Helper
Gold Member
You've done nothing wrong . 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 .

Ok thanks a lot! :D