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Three dimensional vector help

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data


    vector A=(3i+2j+k)m. The angle vector makes with the y axis is nearest
    a 52 degrees
    b 54 degrees
    c 56 degrees
    d 58 degrees
    e 60 degrees

    2. Relevant equations



    3. The attempt at a solution

    I did arctan of (2/3) to get 34 then subtracted it from 90 and got C but I'm not sure about this one the third component is kind of screwing with me a little can anyone clear up how to do this with a third dimension.
     
  2. jcsd
  3. Oct 3, 2007 #2
    for this you need to use the geometric definition of the dot product:

    [tex]\vec{a}\cdot\vec{b}=\parallel\vec{a}\parallel\parallel\vec{b}\parallel \cos \theta[/tex]

    where ||a|| is the distance of a to the origin (0,0,0).
    [tex]\vec{a}[/tex]=[x,y,z]
    [tex]\parallel\vec{a}\parallel=\sqrt{x^2+y^2+z^2}[/tex]

    and theta is the angle between the 2 vectors.

    solving for theta gives you:

    [tex]\theta=arccos\{\frac{\vec{a}\cdot\vec{b}}{\parallel\vec{a}\parallel\parallel\vec{b}\parallel}\}[/tex]

    you know A in vector form and you just have to find the vector form of the y-axis and do the dot product. find the distace's and the arccos
     
  4. Oct 3, 2007 #3
    is the vector component form of the y axis 0,1,0
     
    Last edited: Oct 3, 2007
  5. Oct 3, 2007 #4
    I got arccos ( 2/ sqrt(14)) = 57.7 degrees this is with assuming the vector form of the y axis is 0,1,0
     
  6. Oct 3, 2007 #5
    yes for the y-axis you would use the unit vector j which has coordinates of (0,1,0).
     
  7. Oct 3, 2007 #6
    thanks alot
     
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