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Calculate absolute value of A+B

  1. Jan 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider two vectors by A=5i-3j and B=-i - 2j
    Calculate A+B
    Calculate A-B
    Calculate absolute value of A+B
    Calculate absolute value of A-B


    2. Relevant equations
    R= (Ax + Bx)i + (Ay + By)j
    R= (Ax + Bx)i - (Ay + By)j


    3. The attempt at a solution

    I'm not sure where to start with this problem. I know the equations to use. But its figuring out what value goes where that is hindering me:confused:
     
  2. jcsd
  3. Jan 29, 2007 #2

    Kurdt

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    For the first two you have written the equations you need almost (The subtraction equation needs modifying). What is stopping you plugging in Ax and Bx etc. For parts 3 and 4 do you know how to calculate the absolute value of a vector?
     
    Last edited: Jan 29, 2007
  4. Jan 29, 2007 #3
    still working

    I found the solution to part A. For part B I am putting the numbers into the equation B+(-A). It looks like this (-1-5)+(-2+3) and I get -6i+1j. However, that appears to be wrong. I don't know how to do the absolute value of a vector. And I also don't know how to find the direction of A+B and A-B.
     
  5. Jan 29, 2007 #4

    Kurdt

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    When adding vectors we have as you have above:

    [tex] \mathbf{C}=(A_x +B_x) \mathbf{\hat{i}} + (A_y +B_y)\mathbf{\hat{j}} [/tex]

    For vector subtraction or vector difference we have:

    [tex] \mathbf{C}=(A_x -B_x) \mathbf{\hat{i}} + (A_y -B_y)\mathbf{\hat{j}} [/tex]

    The absolute value is given by pythagorean theorem.

    [tex] |\mathbf{A}| = \sqrt{A_x^2+A_y^2} [/tex]
     
    Last edited: Jan 29, 2007
  6. Jan 29, 2007 #5
    Ok...still a little confused on the absolute value thing.
    For the absolute value of A+B I found the absolute value of A which is square root of 34=5.83. Then found the absolute value of B which is square root of 5=2.236. Then I added those together. I did the same for the absolute value of A-B except i subtracted them instead of adding them. However, this isn't right. Where am I going wrong?

    I am also confused on how to find the direction of A+B and A-B. I thought that you found the direction of something by taking the tangent(-1) of y/x. However, I did that on a problem where y=35.0 and x=-22.5 and got the answer to be -57.3 and this was wrong...Isn't that the correct way though?
     
  7. Jan 29, 2007 #6

    Kurdt

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    With regards to the absolute value, you have done it wrong. What you need to do is take the absolute value of the new vector which is A+B or A-B.

    I'm not sure about your second question because A+B and A-B are vectors they already have direction. Are you talking about finding a specific angle relative to one of the axes?
     
  8. Jan 29, 2007 #7
    So I would take the absolute value of A+B which was 4i+-5j? I would do that by taking the square root of (4squared + -5squared)? And do the same for the absolute value of A-B which was 6i + -1j? Would I subtract those square roots? And the question about their direction wanted to know what there direction would be in degrees from the positive x axis.
     
  9. Jan 29, 2007 #8

    Kurdt

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    That is correct except you would not subtract the square roots for the A-B vector you would add them as normal.

    If you want the angle from the positive x-axis, then taking the inverse tangent of y/x is a method to follow but you have to be careful because you have to add some degrees on depending which quadrant the point is in.
     
  10. Jan 29, 2007 #9
    could you elaborate on finding the degrees from the positive x axis. I am not told what quadrant anything is in...
     
  11. Jan 29, 2007 #10
    I have a similar problem about finding the position from the x-axis. In this problem y is 35.0 and x is -22.5. So wouldn't I take the inverse tangent of 35.0/-22.5? This is my thinking but my answer of -57.3 is incorrect.
     
  12. Jan 29, 2007 #11
    Alright, so I figured out the direction of the A+B and A-B. However, when i do the exact same thing for the second problem where x=-22.5 and y=35.0 it is wrong. Would i not take the inverse tangent of 35.0/-22.5. I find this to be
    -57.26 degrees from the positive x-axis. Why isn't that correct?
     
  13. Jan 29, 2007 #12

    HallsofIvy

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    Are you required to do that? You stated the problem as
    "Consider two vectors by A=5i-3j and B=-i - 2j
    Calculate A+B
    Calculate A-B
    Calculate absolute value of A+B
    Calculate absolute value of A-B"

    None of those require finding a direction.

    However, to answer your question, look at the specific x and y values. x= -22.5 and y= 35.0 so you are in the second quadrant. You -57.3 degrees is from the negative x-axis. The angle from the positive x-axis is 180- 57.3= 122.7 degrees.
     
    Last edited: Jan 29, 2007
  14. Jan 29, 2007 #13
    Yes, I was. However I found the correct solution to that one. However, I have a similar problem that wants to know the degrees from the positive x-axis of a vector with components x=-22.5 and y=35.0. I did the inverse tangent of 35.0/-22.5 and find it to be -57.26 degrees from the positive x-axis. Why isn't that correct?
     
  15. Jan 29, 2007 #14

    Kurdt

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    Thats explained by HallsofIvy above. The angle you calculated was from the negative x-axis not the positive so you need to correct it.
     
  16. Jan 29, 2007 #15
    THANKS!! I think I may finally be getting the hang of this!
     
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