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Angle between radius vector and x,y,z axis

  1. May 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the angle between radius vector of the point M(6,2,9) and:

    a)x axis ; b)y axis ; c)z axis

    2. Relevant equations

    [tex]cos(\vec{a},\vec{b})=\frac{\vec{a} \vec{b}}{|\vec{a}||\vec{b}|}[/tex]

    3. The attempt at a solution

    I tired with

    [tex]cos\alpha=\frac{(6,2,9)(6,0,0)}{\sqrt{121}\sqrt{36}}[/tex]

    but it is not like in my textbook results...
     
  2. jcsd
  3. May 17, 2008 #2

    CompuChip

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    Why don't you switch to polar coordinates? Then you can just read them off.
     
  4. May 17, 2008 #3
    What is the answer it states in the book?
     
  5. May 17, 2008 #4
    [tex]\frac{2*\sqrt{10}}{11}[/tex] for the angle between radius vector and x axis...
     
  6. May 17, 2008 #5
    how about y and z (just confirming my answers)
     
  7. May 17, 2008 #6
    b)y axis

    [tex]\frac{3*\sqrt{13}}{11}[/tex]

    c)z axis

    [tex]\frac{\sqrt{85}}{11}[/tex]

    Please tell me how did you solve this task. Thanks.
     
  8. May 17, 2008 #7

    tiny-tim

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    Hi Physicsissuef! :smile:

    You've got the question wrong …

    … it's the angle betwen the radius vector and the three coordinate planes, not axes.
     
  9. May 17, 2008 #8
    I still don't have the right results...
     
  10. May 17, 2008 #9

    HallsofIvy

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    If <x, y, z> is a unit vector, [itex]cos(\theta_x)= x[/itex], [itex]cos(\theta_y)= y[/itex], and [itex]cos(\theta_z)= z[/itex] give the angles it makes with the axes. In other words, divide <6, 2, 9> by its length.
     
  11. May 17, 2008 #10
    Yes but I didn't get the right answers.

    My answers:

    [tex]\frac{6}{11} ; \frac{2}{11} ; \frac{9}{11}[/tex]

    and their answers are above...
     
  12. May 17, 2008 #11

    tiny-tim

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    Draw a diagram … follow what I said …
    … and use Pythagoras! :smile:
     
  13. May 17, 2008 #12
    I get

    [tex]cos \alpha=\frac{(6,2,9)(6,2,0)}{11*\sqrt{40}}=\frac{40*\sqrt{40}}{11}[/tex]
     
  14. May 17, 2008 #13

    tiny-tim

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    Hi Physicsissuef! :smile:

    No, you get:

    [tex]\cos\alpha=\frac{(6,2,9)(6,2,0)}{11*\sqrt{40}}=\frac{40}{11\sqrt{40}}=\frac{\sqrt{40}}{11}=\frac{2\sqrt{10}}{11}\,.[/tex] :smile:
     
  15. May 17, 2008 #14
    Ok, thanks... Probably there is mistake in my text book. It clearly says that they are axis and not planes. Thanks again...
     
  16. May 17, 2008 #15
    And what about this task?

    Find the angles of the directions of the normal line of the plane 2x-y+2z+9=0

    Probably the normal is [tex]n(2,-1,2)[/tex], probably the directions are the x,y,z axis.

    [tex]cos\alpha=\frac{(2,-1,2)(2,0,0)}{3*2}=\frac{2}{3}[/tex]

    and in my book says it is [itex]cos\alpha=\frac{-2}{3}[/tex]

    for the second it says that it is [itex]cos\beta=\frac{1}{3}[/itex]

    and the third [itex]cos\gamma=\frac{-2}{3}[/itex], what is the problem?
     
  17. May 17, 2008 #16

    HallsofIvy

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    Those are the cosines of the angles the line makes with the coordinate axes.

    Now, solve [itex]cos(\alpha)= 6/11[/itex], [itex] cos(\beta)= 2/11[/itex], and [itex]cos(\gamma)= 9/11[/itex].

    I have no idea why tiny tim is assuming "angles made with planes". A line or vector does NOT make a single angle with a plane. We can interpret "angle with normal to plane" as an "angle with plane" but that puts you back to "angle with coordinate axes" again.
     
  18. May 17, 2008 #17
    And what about my last problem?
     
  19. May 17, 2008 #18

    tiny-tim

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    Hi HallsofIvy! :smile:

    hmm … never could think three-dimensionally …

    but my method seemed to get the right result! :biggrin:

    I've now noticed that the book answers are the sines of the angles with the axes, while 6/11 2/11 and 9/11 are the cosines.

    I'm perplexed as to why the book would expect sines as answers … so I still think it's the planes not the axes. :rolleyes:

    After all, one often sees "a pole makes an angle x with the ground". :smile:
     
  20. May 17, 2008 #19

    tiny-tim

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    You get cos = 2/3, and the books says cos = -2/3.

    Yours is the relative angle "the short way round", and the other is "the long way round".

    wikipedia, at http://en.wikipedia.org/wiki/Euler_angles says:
    So you'll have to check which system the book is using.

    I hope that if it comes up in the exam, the question will tell you which system to use. Is that right, guys?
     
  21. May 18, 2008 #20
    The book is using:

    [tex]cos\alpha=\frac{\vec{n_1}\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}[/tex] for vectors
    and
    [tex]cos\alpha=\frac{|\vec{n_1}\vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}[/tex]
    for lines... Anyways if I use the both formulas I get positive number, I don't know what is the problem...
     
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