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3 vectors

  1. Feb 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Vector A=(0i+3j), vector B=(8i+-1j), vector C=(-8i+5j) Use the component method to determine the following:

    1. the magnitude and direction of Vector D=Vector A+vector B+vector C
    2. vectorE=-A-B+C

    2. Relevant equations



    3. The attempt at a solution

    1. magnitude D=A+B+C=square root (0+8+-8)^2+(3+-1+5)^2=9
    direction=tan-1(3-1+5/0+8-8) this won't work because I am dividing by zero, but I'm not sure what it would be.

    2. Magnitude -A-B+C=square root (-0-8+-8)^2+(-3--1+5)^2=2.65

    direction=tan-1(3+-1+5/0+8-8)=undefined

    Could someone please show me what I'm doing wrong?

    Thank you very much
     
  2. jcsd
  3. Feb 14, 2008 #2
    I did a quick check of your first answer, but it's wrong. You need to recheck your math.

    For the i's (0+8-8) = 0
    For the j's (3-1+5) = 7

    So the magnitude of A+B+C = 7

    Now, since the i's represent the x value, and the j's represent the y value, what do you know about the angle that has an x value of 0, and a y value of 7?
     
  4. Feb 14, 2008 #3
    Thank you very much

    Wouldn't the angle just be 90 degrees? Could you show me what I did wrong with the subtraction one?

    Thank you
     
  5. Feb 14, 2008 #4
    Yes, your angle would just be 90 degrees, or the positive y axis.

    For the second problem, I think you're just working too fast.

    For the i's (-0-8-8) = -16
    For the j's (-3--1+5) = 3

    So the magnitude is [tex]\sqrt{(-16)^2 + 3^2}[/tex]
    which equals 16.28

    I'm going to let you figure out the angle. Just be careful about where it is, because the i value is negative, and the j value is positive.
     
  6. Feb 14, 2008 #5
    Thank you very much

    Is it tan-1(3+1+5/-0-8-8)=-29.4

    180-29.4=150.6?

    Thank you
     
  7. Feb 14, 2008 #6
    Not quite, but close. Be careful with the negative signs. The 3 on should be negative, so it's tan-1(3/-16) etc.
     
  8. Feb 14, 2008 #7
    Thank you very much

    Does this look correct?

    tan-1(3-1-5/-0-8-8)
    =10.62

    Do I then need to subtract it from 180?

    180-10.62=
    169.4

    Thank you
     
  9. Feb 14, 2008 #8
    If you want to show the angle from the x axis, then yes, subtract it from 180. And the answer looks good to me.

    You're very welcome.
     
  10. Feb 14, 2008 #9
    Thank you very much again

    Regards
     
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