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Another Quick Vector Question

  1. Aug 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the vector with the components: Ax=3.2 Ay= -5.15

    3. The attempt at a solution

    -5.15^2 + 3.2^2= 36.76
    sq root of 36.76= 6.06
    So, the magnitude is 6.06. Now for the direction....

    CosD=3.2/6.06
    ArcCos(0.52)=58.7 degrees
    So, the direction is 58.7

    Am I right? Thanks so much!
     
  2. jcsd
  3. Aug 9, 2008 #2
    with y negative and x positive, the vector falls in fourth quadrant. But 58.7 is in first..

    other than that, everything else seems good.

    try using tan theta = y/x .. it's easier to use
     
  4. Aug 9, 2008 #3
    So does that mean...

    Tan theta=-5.15/3.2
    ArcTan(-1.6)= -57.99 degrees

    How can I have a negative angle?
     
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