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Homework Help: Angle Bisectors of Vectors

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data

    Show that vector C = (BA + AB) / (A + B) is an angle bisector of A and B. Where vectors are represented by bold font, and magnitudes are regular font.

    2. Relevant equations

    A ⋅ C = A C cos(θ) ⇒ cos(θ) = (A ⋅ C) / (A C)

    B ⋅ C = B C cos(θ) ⇒ cos(θ) = (B ⋅ C) / (B C)

    3. The attempt at a solution

    We know that if C is a bisector of A and B, then ∠AC =∠BC = θ must be true.

    I set the above equations equal to each other to get;

    (A ⋅ C) / (A C) = (B ⋅ C) / (B C)

    I notice the magnitude C cancels and then cross multiply the expression to get;

    (A ⋅ C)B = (B ⋅ C)A

    I bring the right side over and use identities of dot products to get;

    C ⋅ [BA - AB] = 0

    This is where I am stuck I don't know how to take it any further. I would appreciate a push in the right direction.
    Last edited: Jan 29, 2016
  2. jcsd
  3. Jan 29, 2016 #2


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    Hello Jake, :welcome:

    Fill in something for C to proceed !
    (for example (BA + AB) / (A + B) :smile: )
  4. Jan 29, 2016 #3
    Hi BvU,

    Thanks for responding.

    I don't understand though. How can I fill in something for C. C is some vector C(x,y,z) that bisects A and B. If I simply substitute the solution for C into C, then I will not have proved the expression in the problem statement. I have to prove the relationship is valid.
  5. Jan 29, 2016 #4


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    The problem is NOT to prove "the expression in the problem statement". That is given. The problem is to prove that the two angles are equal using that expression.
  6. Jan 29, 2016 #5
    So, how do I do that given what has been done above? What was suggested was to sub in the expression for C. What then? Do i simplify and see if it goes to zero?

    Please excuse any bad spelling. I am on a phone walking thru campus.
  7. Jan 29, 2016 #6


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    You work it out and see that it IS zero.
  8. Jan 29, 2016 #7

    So why didnt you just say that in the first place? Oh wait you did. I hope the humor in that statement was not lost in text. Thank you very much for taking your time to help me. I truly apreciate it.
  9. Jan 29, 2016 #8


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    You're fine and I'm fine too. No worry !
  10. Jan 30, 2016 #9
    I worked it out using my calculator. It is not zero. Did I do something wrong in my earlier steps? If I did I do not see it.
    Last edited: Jan 30, 2016
  11. Jan 30, 2016 #10
    Please, can someone help. This is due on Monday. I have spent hours on this and am stuck.
  12. Jan 31, 2016 #11
    Seriously need help. Due tomorow. I have been going round and round in circles and cant get it.
  13. Jan 31, 2016 #12


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    Just do what @BvU has been saying all along:

    compute C ⋅ [BA - AB] by using the given value of C:
    = (BA + AB) / (A + B)

    You don't need a calculator for that.
    Last edited: Jan 31, 2016
  14. Jan 31, 2016 #13
    When i make the substitution and carry out the dot product the result is as follows;

    1/(A + B) [ B^2Ax^2 - A^2Bx^2 + B^2Ay^2 - A^2By^2 + B^2Az^2 - A^2Bz^2 ]

    This is not zero. I used my calculator to verify my work and i got the same result.
  15. Feb 1, 2016 #14


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    I don't have a clue how you get that result.
    Let's forget the denominator (A+B) for the moment.
    Just focus on X = (BA + AB)⋅(BA - AB).
    The dot product is distributive over addition, so X = B²A⋅A-BAA⋅B+ABB⋅A-A²B⋅B
    Can you further simplify this, using the properties of the dot product (it is commutative, Y⋅Y=Y², ...)?
    Last edited: Feb 1, 2016
  16. Feb 1, 2016 #15
    Your the man samy. I see it. I did not know Y⋅Y=Y². Clearly i need more work on this. Thank you so much. You really helped me out here.
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