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Homework Help: Finding the Magnitude of a Diplacement Vector

  1. Dec 22, 2009 #1
    1. Cavers spelunked 2.6 km westward, 3.9 southward, and 25 meters upward. What is their displacement vector?

    2. Relevant equations

    3. The attempt at a solution - The formula for the magnitude of a vector which is

    sqrt of [(3.6)^2+(2.9)^2+(0.025)^2] ≈4.623 km
  2. jcsd
  3. Dec 22, 2009 #2
    Let the point P1 lie at (-2600, -3900, 25).

    Then OP1 will be a line segement whos magnitude is denoted as follows

    sqr[ (-2600)^2 + (-3900)^2 + (25)^2 ] = 4687.28m = 4.69km

    Looks about right, assuming it's 3.9km southward.

    EDIT: I failed to read the question, The displacement vector would simply be r = -2600i -3900j + 25k, where i, j, k are the unit vectors of x, y, and z.

    2nd EDIT: I reread the title and if you're looking for the magnitude of that displacement vector it is indeed 4.69km, sorry for the confusion.
    Last edited: Dec 22, 2009
  4. Dec 22, 2009 #3
    Alright, then I am not understanding how to find a displacement vector. I thought that the displacement of a vector was the same as its magnitude. How is finding the displacement of a vector somehow different than finding the magnitude of a vector?
  5. Dec 22, 2009 #4
    The displacement vector conveys both magnitude and direction. You have the distance, does the problem want direction as well?
  6. Dec 22, 2009 #5
    I think you are mixing up terms, its not a vector with a displacement, its a displacement vector, this is a vector that defines displacement of an object based on another parameter(usually time).
  7. Dec 24, 2009 #6
    You were exactly correct. That is what I was doing. Thanks for the replies everyone.
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