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Math Help - Vector proof

  1. #1
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    Vector proof

    Prove that, if (c-b).a = 0 and (c-a).b = 0, then (b-a).c=0. Show that this can be used to prove the following geometrical results.
    (a) The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.

    I can do the first part by using the distributive rule and communitative rule for scalar products.
    But part (a) took a too much effort for what I think is meant to be a simple question. I showed that the lines through the vertices can be expressed in the one vector, therefore they must meet in a point. I used the fact that their dot product is zero. It took a hell of a lot of tedious algebra and I'm not even sure that its right. I'd appreciate any help.

    Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by woollybull View Post
    Prove that, if (c-b).a = 0 and (c-a).b = 0, then (b-a).c=0. Show that this can be used to prove the following geometrical results.
    (a) The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.

    I can do the first part by using the distributive rule and communitative rule for scalar products.
    But part (a) took a too much effort for what I think is meant to be a simple question. I showed that the lines through the vertices can be expressed in the one vector, therefore they must meet in a point. I used the fact that their dot product is zero. It took a hell of a lot of tedious algebra and I'm not even sure that its right. I'd appreciate any help.

    Thanks.
    (\vec c -\vec b) \cdot \vec a=0 and (\vec c - \vec a)\cdot b =0

    Using the distributive law of the dot product we get

    \vec c \cdot \vec a -\vec b \cdot \vec a=0 and \vec c \cdot \vec b -\vec a \cdot \vec b =0

    using the communitive property of dot products we can rewrite the first equation as

    \vec c \cdot \vec a -\vec a \cdot \vec b=0

    Then multiply it by -1 and add it to the 2nd equation

    -\vec c \cdot \vec a +\vec a \cdot \vec b=0
    \vec c \cdot \vec b -\vec a \cdot \vec b =0

    \vec c \cdot \vec b - \vec c \cdot \vec a =0

    Now use the distribitive property in reverse(aka factor) out c

    \vec c \cdot (\vec b-\vec a)=0

    By the communitive property of the dot product we get


     (\vec b-\vec a) \cdot \vec c=0

    Yeah!!
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  3. #3
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    Thanks, but the initial part is fine, it is part (a) which I find to be very messy.
    Does anyone have some advice for the second part of the question.
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