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Thread: 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 $\displaystyle z_a,\ z_b\ and\ z_c$

    $\displaystyle z_d\ =\ ?$

    (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 $\displaystyle z_a,\ z_b\ and\ z_c$

    $\displaystyle z_d\ =\ ?$

    (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 $\displaystyle AB$ onto $\displaystyle AC$. In vector terms we shall find the projection of $\displaystyle z_b-z_a$ onto $\displaystyle z_c - z_a$.

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

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

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

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

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