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The magnitude of the vector difference , is closest to

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data
    The components of vectors and are given as follows:

    Ax = +5.7 Bx = -9.8
    Ay = -3.6 By = -6.5

    The magnitude of the vector difference ,B-A is closest to:

    2. Relevant equations
    A=square root (ax)^2+ay^2

    B= square root (bx)^2 + by^2

    3. The attempt at a solution

    Pretty much what I did was get the magnitude for A by using the above formula, and for A I got 6.7 and for B I got 11.79, I then subtracted B-A to get 5.0. However the answer is 16 I dont understand how that can be the case.

    Thank you
  2. jcsd
  3. Jan 14, 2012 #2

    Simon Bridge

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    Science Advisor
    Homework Helper

    You forgot to account for the direction of the vectors.
    To get the difference between two vectors, you subtract the vectors head-to-tail.

    In terms of components:
    D = B - A = (Bxi + Byj) - (Axi + Ayj)

    what is |D|?
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