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Vectors application Question

  1. Nov 5, 2012 #1
    Hi,

    So, there are three towns: A, B and X

    Town A is 240 km East and 70 km North of O.
    Town B is 480 km East and 250 km North of O.
    Town X is 339 km East and 238 km North of O.

    At A, the airplane changes direction so it now flies towards B. Point Y in the path AB is the closest the airplane ever is to town X.

    They then ask us to show that AB is perpendicular to -3i + 4j, which I can do.

    They then ask us to find Distance XY.

    To do so, I did the following:

    r = (240 + 240t) + (140+280t)

    r = (339 - 3s) + (238 + 4s)

    I then solved to find the point of intersection, which should, supposedly, yield the position of Y.

    I got t as 0.46, meaning Y would be 350.4i + 222.8j.

    I then went on to find the vector XY and work out its magnitude, which gave me 19 km.

    The book, however, claims it is 75km.

    Can anyone help me spot where I went wrong?

    Thanks in advance!
     
  2. jcsd
  3. Nov 5, 2012 #2

    Mark44

    Staff: Mentor

    What is your reasoning for the above?
     
  4. Nov 5, 2012 #3
    Sorry, I meant:

    r= (240+240t) + (140+180t).

    The calculations were performed based on 180 not 280t
     
  5. Nov 5, 2012 #4

    Mark44

    Staff: Mentor

    What I'm asking is, where does this equation come from, especially the 240t and 180t terms?
     
  6. Nov 5, 2012 #5
    Sorry, I got my mistake now. The method was right, I just had gotten the numbers wrong again.

    I'll explain that equation (which should read (240 + 240t)+(70+180t)

    It is the equation for the line AB. A point on the line (A) is 240i + 70j whereas the direction, AB, is 240i + 180j
     
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