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Find the angles of the triangle

  1. Aug 18, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let a be a unit vector and b be a non-zero vector not parallel to a. Find the angles of the triangle, whose two sides are represented by the vectors √3(a x b) and b-(a.b)a

    2. Relevant equations

    3. The attempt at a solution
    The third side will be equal to [itex]\sqrt{3}(a \times b) - \left\{ b-(a.b)a \right\}[/itex]

    Now taking the dot product of the given two sides will give me zero so the angle between them is 90°. But what about the other two angles? Even if I take the dot product of the remaining sides pairwise it will not give me an expression which can be evaluated as I don't know the lengths of sides of the triangle.
     
  2. jcsd
  3. Aug 18, 2013 #2

    mfb

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    You know the length of the sides in terms of a and b, that is sufficient to find an expression for the angle.
    I did not calculate it, but my intuition is that you will get a result independent of a and b.
     
  4. Aug 18, 2013 #3

    haruspex

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    I don't think that is borne out, but does it matter? Why cannot the answer be in terms of a and b?
     
  5. Aug 19, 2013 #4

    ehild

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    It can, but a is unit vector and the result will contain b in the form of b2/b2 :devil:

    ehild
     
  6. Aug 19, 2013 #5

    haruspex

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    OK, I see it now. Must have made an algebraic error. b - (a.b)a is merely the component of b perpendicular to a. The vector axb only depends on that component of b.
     
  7. Aug 19, 2013 #6

    mfb

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    Exactly, and therefore |b| does not influence the angles. |a| is known, and the other lengths scale properly with the angle between a and b, so the result is independent of the vectors.
     
  8. Aug 19, 2013 #7

    utkarshakash

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    Can you please help me out? I'm still not getting it.
     
  9. Aug 19, 2013 #8

    mfb

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    Did you calculate that?
     
  10. Aug 20, 2013 #9

    utkarshakash

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    No. I mean I don't know how to calculate that? My only guess is length of one side of the triangle might be √3. I'm not sure. I have no idea about length of other two sides.
     
  11. Aug 20, 2013 #10

    haruspex

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    Please post your attempt at calculating the angle between, say, [itex]\sqrt{3}(a \times b) - \left\{ b-(a.b)a \right\}[/itex] and [itex]\left\{ b-(a.b)a \right\}[/itex]
     
  12. Aug 20, 2013 #11

    utkarshakash

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    OK. I took the dot product of the two sides (posted by you) and got something like this

    [itex] \left( \vec{a} . \vec{b} \right) ^2 - | \vec{b} |^2 [/itex]

    which after further simplification yields [itex]-b^2 sin^2 \theta [/itex], where θ is the angle between a and b.

    Now, to get the angle between these two sides I also need their magnitudes so that I can divide the obtained expression by product of their magnitudes to get the cosine of the angle between them.
     
    Last edited: Aug 20, 2013
  13. Aug 20, 2013 #12

    mfb

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    Just continue with the calculations...
    You need the magnitudes, how do you calculate them?
     
  14. Aug 21, 2013 #13

    utkarshakash

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    I took the dot product of the sides with itself and got the magnitudes as [itex]4b^2 sin^2 \theta [/itex] and [itex] b^2 sin^2 \theta[/itex]. Now if I will divide the expression obtained earlier with the product of these two I get [itex]\dfrac{-1}{4 b^2 sin^2 \theta}[/itex]. Unfortunately, this still contains b. :cry:
     
  15. Aug 21, 2013 #14

    mfb

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    I think your values are the squares of the magnitudes and not the actual magnitudes.
     
  16. Aug 21, 2013 #15

    utkarshakash

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    Oh! That was a silly mistake. Thanks for pointing out.:approve:
     
  17. Aug 21, 2013 #16
    How do you evaluate ##(a \times b).(a \times b)##?

    Sorry for the dumb question but how do you conclude that? :confused:
     
  18. Aug 21, 2013 #17

    haruspex

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    Useful rule: ##(a \times b).(a \times b) = a^2b^2sin^2(\theta) = a^2b^2(1-cos^2(\theta)) = a^2b^2-(a.b)^2##.
    Since a is a unit vector, (a.b)a is the component of b in the direction of a. Its magnitude is |b|cos(θ) and its direction is that of a. Subtracting it from b leaves the component perpendicular to a.
     
  19. Aug 22, 2013 #18
    Thanks haruspex. :)
     
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