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Adding the vectors ijk notation

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Add the following 3 displacement vectors and give the answer in ijk notation and in magnitude and direction format.
    A = 3i - 3j m
    B = i - 4j m
    C = -2i + 5j m.

    Then I have to do a scetch of the resultant vector...which i'll call D.

    2. Relevant equations

    3. The attempt at a solution
    So I know to get vector D's i hat and j hat components...i just add the i's and j's of the vectors that make it up so

    i = 3 + 1 - 2 = 2
    j = -3 - 4 + 5 = -2


    so then to find the magnitude of D i do

    √(2)² + (-2)²

    which comes out to √8 which is roughly 2.83 m here

    Then to find the angle i do the inverse tan of (-2/2) or -1....which comes out to -45°

    Now to graph....should I find the magnitude and angle of each vector?

    A = √(3)² + (-3)² = √18 = 4.24 m at -45°
    B = √(1)² + (-4)² = √17 = 4.12 m at -76°
    C= √(-2)² + (5)² = √29 = 5.39 m at -68°

    Do I graph them just like this? Because how can the resultant vector be at the SAME angle at vector A? It would be a smaller version of vector A because the magnitude of A is >D and the angle is the same.

    Can anyone point out where I'm misunderstanding this?
  2. jcsd
  3. Feb 17, 2013 #2
    The angle of the resultant vector is not 45°. In which quadrant of the XY plane is it?

    The problem with arctan is that it confuses (-a)/b with a/(-b), and (-a)/(-b) with a/b. So you have to keep an eye on the signs to determine which quadrant it is, and what the angle really is.
  4. Feb 17, 2013 #3
    You don't need to work out the magnitude and angle of the vectors.
    For example, for A = (3i - 3j) m, the x-component is 3m and the y-component is -3m.
    So, just draw a line from the origin to the point (3m, -3m)
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