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Vector which has same angle with x,y,z axes

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data?
    How do i find a vector that has same angle with the three coordinate axes (x,y,z)?

    3. The attempt at a solution

    I immediately thought [1,1,1] would be it but it's not. I'm trying to find a plane whose normal vector forms the same angle with the three coordinate axes.
     
  2. jcsd
  3. Aug 23, 2012 #2

    Bacle2

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    Why doesn't v=(1,1,1) work?

    Let e,i,j stand for (1,0,0) ,(0,1,0) and (0,0,1) respectively.

    Using a.b=|a||b|cost,

    Then v.e=v.i=v.j =1 ; |e|=|i|=|j|=1 .Then

    v.e/|v||e|= v.i/|v|.|i|= v.j/|v||j|=cost

    Or did you have a different notion of angle in mind?
     
    Last edited: Aug 23, 2012
  4. Aug 23, 2012 #3
    hmm...it does work....

    ok i need to ask a follow up question but i'll create a new thread
     
  5. Aug 23, 2012 #4

    HallsofIvy

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    If [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex] are three angles, the unit vector that makes angle [itex]\alpha[/itex] with the x-axis, angle [itex]\beta[/itex] with the y-axis and [itex]\gamma[/itex] with the z-axis is [itex]cos(\alpha)\vec{i}+ cos(\beta)\vec{j}+ cos(\gamma)\vec{k}[/itex]. If all angles are the same, those three cosines are the same so any vector of the form (x, x, x), and in particular (1, 1, 1) will make equal angles with the three coordinate axes.

    perhaps you are looking for the unit vector. The length of (x, x, x) is [itex]\sqrt{x^2+ x^2+ x^2}= x\sqrt{3}[/itex] and we want that equal to 1: we want [itex]x= 1/\sqrt{3}= \sqrt{3}/3[/itex]. The unit vector that makes equal angles with the coordinate axes is [itex](\sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3)[/itex].
     
  6. Aug 23, 2012 #5
    yes i got the same result for the unit vector. Thanks Ivy. Please see my other thread on the stress tensor. its a follow up to this question
     
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