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Find the angle between the two vectors

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data
    What is the angle between vectors E and F in the diagram?
    Use components to determine the magnitude and direction of G= E+F


    2. Relevant equations

    sqrt(x^2 + y^2) is the equation for magnitude


    3. The attempt at a solution

    All right. Given by the diagram that the end point of F is (-1,-2), I calculated that F's magnitude would be sqrt(5). Given E's diagram, its end point is (1,1), so its magnitude was calculated to be sqrt(2).

    Then since I have opposite and adjacent calculations, I would use tangent-
    tan (theta) = sqrt(2)/sqrt(5), and then take the arc tangent of that to get an answer of 35.9...which is wrong. Why?

    In regard to the second question about G, I am stumped. How do you add vectors together? Do you just add their coordinates?
     

    Attached Files:

  2. jcsd
  3. Aug 28, 2011 #2

    rock.freak667

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    F = (-1,2)

    Yes you would add together the corresponding components.
     
  4. Aug 28, 2011 #3
    Whoops. But either way, that doesn't change the fact that it turns out to be sqrt(5), and I'm still getting the wrong answer....
     
  5. Aug 28, 2011 #4

    Well that went well, I got the magnitude correct as 3.00, but then how would I figure out the direction of G? All I know is G's end point.
     
  6. Aug 28, 2011 #5

    rock.freak667

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    Yes but you need to add together the angles between the individual vectors and the y-axis. By chance, do you know how to calculate the dot product of two vectors?

    Well what is the end point of G?
     
  7. Aug 28, 2011 #6
    Why would I have to do vectors F and E separately? Can't I just draw a line between them, call that the hypotenuse, and use my magnitudes that I found? I don't even know what a dot product is... so I guess that answers your question!!

    The end point of G is (0,3), which I would think would give me a right angle from the origin, but that's definitely not the right answer.
     
  8. Aug 28, 2011 #7

    PeterO

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    A right angle would be fine if you are considering the angle with the positive x axis.
     
  9. Aug 28, 2011 #8
    All right, that was correct, and that makes sense because its coordinates were directly up the y-axis. But what of the two vectors that I have to find the angle of? Why is arctan not working for my values?
     
  10. Aug 28, 2011 #9

    PeterO

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    That 35.9 degrees you got may be tha angle between the y-axis and the vector, so you have to add an extra 90 to get the base reference.

    I seem to recall we base vectors on polar co-ordinates, so there is just a single ray for the measurement [like the positive x-axis] and an angle from that ray, with anticlockwise taken as positive.
     
  11. Jan 29, 2013 #10
    don't mean to necro this thread, but i am very confused on how to find the angle between the vectors (phi). how do u do that?
     
  12. Jan 29, 2013 #11
    tangent of theta is not sqrt(2)/sqrt(5). For this to be correct, these must be the opposite and adjacent sides of the same right triangle. I don't see any such right triangle in the figure. What is the angle between E and the y axis? What is the angle between F and the y axis. The sum of these two angles is the angle between E and F.
     
  13. Jan 29, 2013 #12
    all that was given was in the picture attached to the OP's post.
    Only the magnitudes could be calculated. nothing was really given
     
  14. Jan 29, 2013 #13

    PeterO

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    Just as you could use the co-ordinates on the original figure to get the magnitude of each vector, you could use those same co-ordinates to calculate the angles mentioned by "Chestermiller".
    Actually you should be able to recognise the size of one of the angles - as you will (should) have come across it so many times before.
     
  15. Jan 30, 2013 #14

    PeterO

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    I hope you have appreciated that finding the angle between the two vectors and the magnitude of the sum of the vectors are two quite independent questions.
     
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