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Math Help - Complicated Vector Problem.

  1. #1
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    Thumbs up [SOLVED] Complicated Vector Problem.

    The Problem is:

    "A horseman left the village A at point (1,4) and began riding along a straight road whose direction was given by the vector v=7i+4j. The at some point he turned at a right angle; He never changed direction again until he arrived in the village [LaTeX ERROR: Convert failed] at the point with the coordinates (8,13). At the point where he made the turn he buried a jar full of silver coins, Unfortunately, he forgot the coordinates of the point. Find the point.

    So I drew the graph and labeled the point where he changed direction as C with the coordinates (x,y). This means the vector CA is <x-1, y-4> and the vector BC is <8-x,13-y>.

    Next, I worked out the dot product of the two vectors to be (x-1)(8-x)+(y-4)(13-y)=0 It has to be equal to 0 since the angle is a right angle and it is orthogonal. Been working on this problem for about 45 minutes, and have no idea how to solve it. Any help is appreciated.
    Last edited by Bracketology; August 30th 2011 at 07:26 PM. Reason: SOLVED
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  2. #2
    Junior Member bondesan's Avatar
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    Re: Complicated Vector Problem.

    Think in a triangle. Try doing the problem in terms of projections. You can tell that the vector AX ( X=(x,y) the unknown point) is the projection of AB onto the direction vector v=\{7,4\} (since there is a right angle).
    So you write: AX=proj_{\{7,4\}}AB. Then you have AX=\left(\frac{\{7,9\}\cdot \{7,4\}}{\{7,4\}\cdot \{7,4\}}\right)\{7,4\}.
    Solving this results in AX = \dfrac{17}{13}\cdot\{7,4\}=\left\{\dfrac{119}{13}, \dfrac{68}{13} \right\}.
    As AX=X-A\Rightarrow X=\left\{\dfrac{119}{13}, \dfrac{68}{13} \right\}+\{1,4\}=\left\{\dfrac{132}{13}, \dfrac{120}{13} \right\}.

    Now you could use the fact that AX\cdot XB=0 to check if the answer is correct.
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  3. #3
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    Re: Complicated Vector Problem.

    Thanks! I didn't think about it in terms of projections. This helps a lot.
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