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Math Help - Complex geometry

  1. #1
    MHF Contributor alexmahone's Avatar
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    Complex geometry

    Given triangle ABC, D is the foot of the perpendicular from A to BC. Find the coordinates of D as a complex coordinate.

    Given z_a,\ z_b\ and\ z_c

    z_d\ =\ ?

    (I need this to solve a bigger problem.)
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  2. #2
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    Quote Originally Posted by alexmahone View Post
    Given triangle ABC, D is the foot of the perpendicular from A to BC. Find the coordinates of D as a complex coordinate.

    Given z_a,\ z_b\ and\ z_c

    z_d\ =\ ?

    (I need this to solve a bigger problem.)
    Remember you can think of complex numbers are vectors too. What we shall do is project the line segment AB onto AC. In vector terms we shall find the projection of z_b-z_a onto z_c - z_a.

    Remember \text{proj}_{\bold{v}}(\bold{u}) = \frac{\bold{u}\cdot \bold{v}}{\bold{v}\cdot \bold{v}} and the dot product for complex numbers is \tfrac{1}{2}(z_1\bar z_2 + \bar z_1 z_2).

    Therefore, the projection is, \frac{(z_b - z_a)\cdot (z_c - z_a)}{(z_c - z_a)\cdot (z_c - z_a)} = \frac{(z_b - z_a)(\overline{ z_c -z_a}) + (\overline{z_b-z_a})(z_c - z_a)}{2|z_c-z_a|^2}

    This simplifies to, k=\frac{\Re (z_bz_c) - \Im (z_az_b + z_az_c) - |z_a|^2}{|z_c-z_a|^2} (which is of course a real number).

    Therefore, k(z_c-z_a) is the projected vector.

    Therefore, z_d = z_a + k(z_c-z_a).
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