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Vector Proof

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    If
    c=|a|b+|b|a where a,b, and c are all non zero vectors, show that c bisects the angle between a and b


    2. Relevant equations



    3. The attempt at a solution

    I'm taking the approach to prove that the angle between b and c= the angle between c and a

    I have written the fact that a[itex]\bullet[/itex]b=|a||b|cos [itex]\theta[/itex]
    However, I'm not sure how I can work the equation for c to be able to apply the identity.
     
  2. jcsd
  3. Sep 14, 2011 #2

    lanedance

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    as you know they are in the same plane and c will be between a and b
    [tex]a \bullet b = |a||b|cos\theta [/tex]
    [tex]c \bullet b = |c||b|cos\phi [/tex]

    and the show [itex]\phi = 2\theta [/itex]
     
  4. Sep 15, 2011 #3
    I'm not quite sure what you are suggesting.

    I can draw the relationship

    |a||b|cos [itex]\theta[/itex] = |c||b| cos 2[itex]\theta[/itex]


    I can eliminate the |b| from both sides, but I don't know where to go from there, since |c| doesn't seem to help when substituting.
     
  5. Sep 15, 2011 #4

    HallsofIvy

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    Then get rid of "c". You are given that c= |a|b+ |b|a so that c.b= (|a|b+ |b|a).b= |a||b|^2+ |b|a.b.
     
  6. Sep 15, 2011 #5
    I can replace the c[itex]\bullet[/itex]b side with what you've suggested but then how am I supposed to include that it is twice the angle. I loose this ability without the trigonometric function.
     
  7. Sep 15, 2011 #6
    Can anyone help me?
     
  8. Sep 16, 2011 #7

    lanedance

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    try doing what Halls suggested and let us see what you get
     
  9. Sep 16, 2011 #8

    vela

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    Is this actually true?
     
  10. Sep 16, 2011 #9

    vela

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    You're not supposed to assume it's twice the angle. That's what you're trying to prove!

    Note that HallsofIvy's suggestion [itex]\vec{c}\cdot\vec{b} = ab^2+b(\vec{a}\cdot\vec{b})[/itex] has the dot product of [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex] in it. That will introduce [itex]\cos\theta[/itex] into the equation.

    You can get rid of [itex]c = \sqrt{\vec{c}\cdot\vec{c}}[/itex] from the righthand side by again using the definition of [itex]\vec{c}[/itex].
     
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