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Angles in 3d

  1. Jan 4, 2012 #1
    Hi guys,

    Wondering if you could help me on this one. If you have a vector in xyz, and you know the angles that the vector is inclined at to two of the axis, how do you find the 3rd one.

    eg, line inclined at 60˚ to the x axis and 45˚ to the y axis, how do you find the inclination to the z axis (which is 60˚ or 120˚ by the way) I know it has something to do with direction ratios and direction cosines, but don't know how to get there. I also know that direction cosines add up to 1, but I can't find a connection.

    Thanks in advance

  2. jcsd
  3. Jan 4, 2012 #2


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    Use "direction cosines". If [itex]\theta[/itex], [itex]\phi[/itex], and [itex]\psi[/itex] are the angles a direction makes with the x, y, and z axes respectively, then [itex]cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}[/itex] is the unit vector in that direction.

    That means that if you are given a vector, v, you can find the angles it makes with the axes by reducing it to a unit vector- divide by its length- and look at the components.

    In your example, you know that a unit vector in your direction is [itex]<cos(60), cos(45), cos(\psi)>= < 0.5, 0.707, cos(\psi)>[/itex] so you must have [itex].25+ 0.5+ cos^2(\psi) = 1[/itex]. You can solve that for [itex]\psi[/itex].
  4. Jan 4, 2012 #3
    Yes of course, that is so simple when it is explained. Thank you very much Hallsofivy.

    Happy new year.

    Rob K
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