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Angle of vector help

  1. Sep 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Vector B has x y and z components of 6.7, 4.7, 2.4 and the magnitude of Vector B is equal to 8.599

    what is the angle between Vector B and the x-axis. answer in units of degrees


    3. The attempt at a solution

    i am not sure where to begin on this one... i hate online physics courses BLAH
     
  2. jcsd
  3. Sep 12, 2007 #2
    can i figure this out wiht a formula or do i need to draw it out?
     
  4. Sep 12, 2007 #3

    Kurdt

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    Have you covered direction cosines of vectors?
     
  5. Sep 12, 2007 #4
    i dont recall covering that
     
  6. Sep 12, 2007 #5
    would i assume 8.599 squared = 6.7squared + A squared
    A = 5.39

    then use those as the lenghts of the sides for the triangle to use the cosine of the angle to get my answer?
     
  7. Sep 12, 2007 #6
    then solving for that angle i get 38.8162364?
     
  8. Sep 12, 2007 #7
    Draw a rectangle with the dimensions of the dimensions (lol). Then use pythagoras on the faces to find the angle.

    That's THE best way to deal with 3D vectors.
     
  9. Sep 12, 2007 #8

    Kurdt

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    Ok. Consider a vector [itex]\mathbf{a}=(a_1, a_2, a_3)[/itex] and imagine the angles between that vector and the vectors [itex]\mathbf{\hat{i}}[/itex], [itex]\mathbf{\hat{j}}[/itex] and [itex]\mathbf{\hat{k}}[/itex] are [itex]\alpha[/itex], [itex]\beta[/itex] and [itex] \gamma [/itex] respectively. Knowing that the magnitude of the vector is [itex]a[/itex] we can use pythagorean theroem to say that:

    [tex] cos(\alpha) =\frac{a_1}{a} [/tex]

    [tex] cos(\beta) = \frac{a_2}{a} [/tex]

    [tex] cos(\gamma) = \frac{a_3}{a} [/tex]

    Thus we note:

    [tex] \mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}=cos(\alpha)\mathbf{\hat{i}}+cos(\beta)\mathbf{\hat{j}}+cos(\gamma)\mathbf{\hat{k}}[/tex]

    Now knowing thiscan you find the angle between the x-axis and the vector with the information you have?
     
    Last edited: Sep 12, 2007
  10. Sep 12, 2007 #9
    i am totally lost on ur idea to draw a rectangle of that... once i draw that rectangle how do i use pythagoras on the faces to find the angle?
     
  11. Sep 12, 2007 #10
    ok KURDT i solved the cos of all 3 of those angles

    the first angle is 38.816 degrees
    the 2nd angle is 56.867 degrees
    the 3rd angle is 73.79337 degrees

    correct???
     
  12. Sep 12, 2007 #11

    Kurdt

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    Yeah they look fine. So its relatively simple once you know about direction cosines which very few establishments teach in my experience.
     
  13. Sep 12, 2007 #12
    im not sure my next step once i have those 3 values
     
  14. Sep 12, 2007 #13
    http://i14.tinypic.com/4y548c8.png

    That's what I mean. If you can visualize the vector, you'll be able to solve the problem faster. In an exam, you don't have time to waste.
     
  15. Sep 12, 2007 #14
    kurdt once i solve for those 3 values what step do i take next??

    i understand why i get those 3 angles im jsut lost on the last steps i need now
     
  16. Sep 12, 2007 #15

    Kurdt

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    You have the answer. The first angle is the angle between the vector and the x-axis. The second is the angle between the vector and the y-axis and the 3rd is the angle between the vector and the z-axis.
     
  17. Sep 12, 2007 #16

    Dick

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    Do you know dot products? That's a much simpler way out of this mess.
     
  18. Sep 12, 2007 #17
    so from the original problem i posted... .where the angle between Vector B with an arrow drawn over the B and x axis is 38.826 degrees??? becuz when i enter that onto the online classroom it is wrong
     
  19. Sep 12, 2007 #18
    I get 35.049. Try that.
     
  20. Sep 12, 2007 #19
    i tried that black wyvern and it was incorrect
     
  21. Sep 12, 2007 #20

    Kurdt

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    Thats because the magnitude you gave seems to be slightly out using the components you gave. Check them and make sure they are correct.
     
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