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Math Help - Vectors, ABC isoceles Triangle.

  1. #1
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    Vectors, ABC isoceles Triangle.

    Consider the triangle ABC. Let P divide the segment AB in the ratio s : t, and Q divide the segment BC in the ratio t : s.



    (a) Draw a sketch of the situation.
    (b) Express the vector PQ in terms of t, s, and AC.
    (c) Hence explain why PQ and AC are parallel




    I understand Part A and realise that the length of AB = PQ, and can also see how PQ is parallel to that of AC. Just unsure on how to express PQ.
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  2. #2
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    Re: Vectors, ABC isoceles Triangle.

    Quote Originally Posted by TehNewbie View Post
    I understand Part A and realise that the length of AB = PQ
    What?

    Quote Originally Posted by TehNewbie View Post
    Just unsure on how to express PQ.
    Let $\lambda=\dfrac{t}{s+t}$. Show that $PB / AB = BQ/BC=\lambda$. Using $\lambda$, express vector PB through AB and BQ through BC. Then represent PQ as PB + BQ and use the fact that AB + BC = AC.
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    Re: Vectors, ABC isoceles Triangle.

    Quote Originally Posted by TehNewbie View Post
    Consider the triangle ABC. Let P divide the segment AB in the ratio s : t, and Q divide the segment BC in the ratio t : s.
    (b) Express the vector PQ in terms of t, s, and AC.
    (c) Hence explain why PQ and AC are parallel
    I will let you make a diagram.
    Note that $\overrightarrow {PQ} = \overrightarrow {PB} + \overrightarrow {BQ} $.

    Now $\overrightarrow {PB} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AB} \, \,,\;\overrightarrow {BQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {BC} \,\& \,\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} $

    You put this together and show us what you find.
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    Re: Vectors, ABC isoceles Triangle.

    Quote Originally Posted by Plato View Post
    I will let you make a diagram.
    Note that $\overrightarrow {PQ} = \overrightarrow {PB} + \overrightarrow {BQ} $.

    Now $\overrightarrow {PB} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AB} \, \,,\;\overrightarrow {BQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {BC} \,\& \,\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} $

    You put this together and show us what you find.
    So does that mean that vector PB = (t/s+t) * vector AC ?

    Edit: Yeap it sure does. Thanks for the help guys I really appreciate it
    Last edited by TehNewbie; April 3rd 2014 at 11:09 PM.
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    Re: Vectors, ABC isoceles Triangle.

    Quote Originally Posted by TehNewbie View Post
    So does that mean that vector PB = (t/s+t) * vector AC ?
    t/(s+t), for goodness' sake!
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    Re: Vectors, ABC isoceles Triangle.

    Quote Originally Posted by TehNewbie View Post
    So does that mean that vector PB = (t/s+t) * vector AC ?
    Actually from $\overrightarrow {PQ} = \overrightarrow {PB} + \overrightarrow {BQ} $.

    and $\overrightarrow {PB} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AB} \, \,,\;\overrightarrow {BQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {BC} \,\& \,\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} $

    You get $\overrightarrow {PQ} = \left( {\frac{t}{{s + t}}} \right)\overrightarrow {AC} $
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    Re: Vectors, ABC isoceles Triangle.

    Quote Originally Posted by emakarov View Post
    t/(s+t), for goodness' sake!
    Haha my bad incorrect notation but you get what I meant :P
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