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Math Help - Dot product question

  1. #1
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    Dot product question

    Consider the triangle ABC in \mathbb{R}^3 formed by the points A(3,2,1), B(4,4,2) and C(6,1,0)

    Find the coordinates of the point D on BC such that AD is perpendicular to BC
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  2. #2
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    Quote Originally Posted by acevipa View Post
    Consider the triangle ABC in \mathbb{R}^3 formed by the points A(3,2,1), B(4,4,2) and C(6,1,0)

    Find the coordinates of the point D on BC such that AD is perpendicular to BC
    If \mathbf{AD} is perpendicular to \mathbf{BC}, then \mathbf{AD}\cdot\mathbf{BC} = 0.

    If D = (x, y, z), then

    \mathbf{AD} = \mathbf{AO} + \mathbf{OD}

     = \mathbf{OD} - \mathbf{OA}

     = (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} + \mathbf{k})

     = (x - 3)\mathbf{i} + (y - 2)\mathbf{j} + (z - 1)\mathbf{k}.



    \mathbf{BC} = \mathbf{BO} + \mathbf{OC}

     = \mathbf{OC} - \mathbf{OB}

     = (6\mathbf{i} + 1\mathbf{j} + 0\mathbf{k}) - (4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})

     = 2\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}.



    So \mathbf{AD}\cdot\mathbf{BC} = 0

    [(x - 3)\mathbf{i} + (y - 2)\mathbf{j} + (z - 1)\mathbf{k}]\cdot[2\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}] = 0

    2(x - 3) - 3(y - 2) -2(z - 1) = 0.


    You are told that (x, y, z) lies on the line BC, so I would advise finding the parametric or Cartestian equations of BC...
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