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Physics I: Vector Subtraction

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data
    C = B - A
    A = 16 Units
    A-theta = 42 degrees from the y-axis
    Ai is pointing in second quadrant

    B = 7 units
    B-theta = 31 degrees from the x-axis
    B is pointing in the third quadrant
    2. Relevant equations


    3. The attempt at a solution
    I am suppose to find the magnitude of C from the components given. When I attempted it, I did it using trigonometric. That got marked wrong. Just a random question, am I able to do it using cross-product or something like that?

    Va-x = -16*sin(42)
    va-y = 16*cos(42)

    Vb-x = - 7.0*sin(31)
    Vb-y = - 7.0cos(31)

    Did Pythagorean theorem and combined the x and y and got the following result:
    C =15.45
     
  2. jcsd
  3. May 12, 2013 #2

    rude man

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    C is also a vector, so you need to come up with an angle as well as the correct magnitude.

    Va_x and Va_y are correct.
    Vb_x and Vb_y are incorrect. 31 deg. is from the x, not the y, axis ...
     
  4. May 13, 2013 #3
    Oops...

    I fixed that. I am getting the correct answer now.
     
  5. May 13, 2013 #4

    SteamKing

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    Vector subtraction has nothing to do with the cross product.
     
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