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Prove this using vectors

  1. Aug 16, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If a,b,c and k are real constants and α,β,γ are variables subject to the condition that atanα+btanβ+ctanγ = k, then prove using vectors that tan^2 α+tan^2 β+ tan^2 γ ≥ k^2/(a^2+b^2+c^2)


    2. Relevant equations

    3. The attempt at a solution
    [itex](ai+bj+ck).(tan \alpha i+ tan \beta j+ tan \gamma k) = k \\
    k^2 = (a^2+b^2+c^2)(tan^2 \alpha + tan^2 \beta + tan^2 \gamma)[/itex]

    But what I arrive at is an equation instead of the inequality required to prove.
     
  2. jcsd
  3. Aug 16, 2013 #2

    ehild

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    Correct so far...
    That is wrong. How is the dot product of two vectors related to their magnitudes?


    ehild
     
  4. Aug 16, 2013 #3

    utkarshakash

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    [itex]\vec{c} ^2 = \vec{c} . \vec{c} = |\vec{c}|^2[/itex]

    I've used this identity to simplify it further.
     
  5. Aug 16, 2013 #4

    ehild

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    Yes, but you have the dot product of two different vectors to be squared. ##(\vec a \cdot\vec b)^2≠|\vec a |^2 |\vec b|^2##.

    ehild
     
  6. Aug 16, 2013 #5

    haruspex

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    That's only the special case where the two vectors are the same. What is the more general relationship?
     
  7. Aug 16, 2013 #6
    Do you know about Cauchy-Schwarz inequality?
     
  8. Aug 16, 2013 #7

    utkarshakash

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    I encountered it in Calculus but never bothered to go through it as it is not in the JEE syllabus.
     
  9. Aug 16, 2013 #8
    Cauchy-Schwarz inequality uses dot product, it isn't too difficult.

    Check out Wikipedia.
     
  10. Aug 17, 2013 #9

    utkarshakash

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    Thanks mate. This inequality helped me to solve some other problems as well.
     
  11. Aug 17, 2013 #10
    Glad to help. :)
     
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