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Thread: triangles represented using vectors

  1. #1
    Apr 2008

    triangles represented using vectors

    Consider a triangle ABC formed by the vectors a=(2,10,25) and b=(5,4,-2) attached at the point A. I've calculated the area of the triangle as $\displaystyle 40.5\sqrt{5} $ and now need to find the vector h=AM, the height of the triangle [so AM is perpendicular to BC and M lies on the line BC]. I've tried using so many different formulae to try and find h or the third side of the triangle and nothing I get seems sensible. Can someone please give me some hints or suggestions on where to start so I can work through this problem? Thanks.
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  2. #2
    MHF Contributor
    Mar 2010
    The line going through $\displaystyle A$ and $\displaystyle B$ is just $\displaystyle A+\gamma(B-A)$, where $\displaystyle \gamma$ is a real number. So $\displaystyle \gamma=0$ gives point $\displaystyle A$ and $\displaystyle \gamma=1$ gives point $\displaystyle B$.

    The altitude $\displaystyle AM$ of the triangle has to be perpendicular to that line, so

    $\displaystyle M\cdot(B-A)=(A+\gamma(B-A))\cdot(B-A)=0$

    which you should be able to solve for $\displaystyle \gamma$, and then find $\displaystyle M$.

    Post again in this thread if you're still having trouble.

    - Hollywood
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