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Three Dimensional Vectors

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Find magnitude and direction

    [tex]\stackrel{\rightarrow}{C}[/tex] = -2i-3j+4k

    2. Relevant equations
    Trig. Functions
    Pythagorean Theorem

    3. The attempt at a solution

    C = [tex]\sqrt{4+9+16}[/tex] = [tex]\sqrt{29}[/tex]

    I don't know what to do for the angles because I'm not sure where the reference point is supposed to be.
  2. jcsd
  3. Sep 15, 2009 #2


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    The magnitude is OK. As far as the "direction" is concerned, you need two angles to specify direction. If you imagine the tip of the vector being on the surface of a sphere (like the Earth), then you need to give the "latitude" angle θ and the "longitude" (measured from the x axis) angle φ.
  4. Sep 15, 2009 #3
    Ah, ok.

    If θ is the latitude one, and A is the angle from the Z-axis to the vector, then:

    θ = 90° - A

    cos(A) = [tex]\sqrt{13/29}[/tex]
    A = cos-1([tex]\sqrt{13/29}[/tex]) = 47.97°

    θ = 42.03°

    If Ø is the longitude one, then:

    Ø = 90 + tan-1(2/4)
    Ø = 116.57° <---- Where I have a problem

    My teacher says that the answer is 112°, but whenever I do it I get 116.57°. What am I missing?
  5. Sep 15, 2009 #4


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    If you project the tip of the vector onto the xy plane, you get the point (-2, -3, 0). Draw a line from the origin to that point. What is the angle it makes with respect to the x-axis?
  6. Sep 16, 2009 #5

    Isn't that the situation though?

    In this case, Ø = 180 - arctan(2/4)
    Last edited: Sep 16, 2009
  7. Sep 17, 2009 #6
    Guys, I need to know if I'm doing this correctly. The homework assignment is due tomorrow.
  8. Sep 17, 2009 #7
    Ok, so that would be arctan (-3/-2) or 56.3 degrees. Which means that the angle from the x-axis would be 236.3 degrees going clockwise or 123.7 degrees going counter clockwise.
  9. Sep 17, 2009 #8


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    Sounds OK. Good luck with your homework.
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