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Find the magnitude and direction of vectorA - vector B?

  1. Feb 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Vector A⃗ has a magnitude of 20m and makes an angle of 30∘ above the positive x axis. Vector B⃗ has a magnitude of15m and is oriented 60∘ to the left of the y axis.
    Find the magnitude of A⃗ −B⃗ .
    Find the direction of A⃗ −B⃗ .
    2. Relevant equations
    sin 150 = sin 30 = 0.5
    cos 150= - cos 30
    Angle of direction = tan-1 (y/x)
    3. The attempt at a solution
    so far
    Ax
    = 20 cos 30 =17.3 m
    Ay = 20 sin 30 = 10 m

    Bx = 15cos 150 = -12.99 m
    By=15 sin 150 = 7.5 m
    add Ax+Bx= 4.31
    Ay+By= 17.5
    sqrt of [(4.31)^2+(17.5)^2]= 18.0
    but i dont know where to go from here ?
     
    Last edited: Feb 7, 2016
  2. jcsd
  3. Feb 7, 2016 #2
    Hi bigcookie:

    This problem statement does not state a problem. What is the question whose answer is sought?

    Regards,
    Buzz
     
  4. Feb 12, 2016 #3
    Hi @bigcookie:
    I see that you edited the problem statement . I apologize for not checking on the thread more promptly.

    I think you have calculated the x y coordinates for A and B correctly, but I don't understand why you ADD the coordinates. The problem statement is interested in A-B.

    Regards,
    Buzz
     
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