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Vectors, Adding, Subtracting.

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Three vectors a , b , and c each have a magnitude of 42 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 29°, 197°, and 314°, respectively. What are (a) the magnitude and (b) the angle of the vector a+ b + c, and (c) the magnitude and (d) the angle of a - b + c? What are (e) the magnitude and (f) the angle of a fourth vector d such that (a + b) - ( c+d ) = 0?

    2. Relevant equations



    3. The attempt at a solution

    I do not how to approach this problem. Could someone please give me a few steps to try. Would be greatly appreciated!
     
  2. jcsd
  3. Sep 13, 2009 #2

    rl.bhat

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    Draw the vectors according their positions. Resolve them along x-axis and y-axis.
    Find the resultant of x and y components. And get the final answer using Pythagoras theorem.
    Whenever the vector is negatived, reverse the vector and proceed.
     
  4. Sep 13, 2009 #3
    I got A=36.73i + 20.36j
    B= -19.07i + (-37.42j)
    C= 29.18i + (-32.21j)

    a+b+c= 46.84i - 29.27j

    It was not correct? Do you see any mistakes?
     
  5. Sep 13, 2009 #4

    Hurkyl

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    We can't see the mistakes if you don't show your work....
     
  6. Sep 13, 2009 #5
    Ok.

    I did 42*cos(29)=36.73
    42*sin(29)=20.36

    42*cos(63) Got the 63 from 260-197...and that =19.07, which i made negative
    42*sin(63)=37.42..also made negative

    and..

    42*cos(46)..Got the 46 from 360-314...and that = 29.18
    42*sin(46)= I just did the calculation, now i'm get 30.21 instead of 32.31....other than that mistake, do you see anything else?
     
  7. Sep 13, 2009 #6

    rl.bhat

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    [42*cos(63) Got the 63 from 260-197...and that =19.07, which i made negative]

    This is wrong. It should be 270 - 197
     
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