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Proof of orthogonality

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex] \Pi \subset \mathbb{R}^3 [/itex] is a plane, and that P is a point not on [itex] \Pi [/itex]. Assume that [itex] Q \in \Pi [/itex] is a point on [itex] \Pi [/itex] whose distance to P is minimal. Show that the vector PQ is orthogonal to [itex] \Pi [/itex]. Define a differentiable vector function r(t) with r(t) [itex] \in \Pi [/itex] and r(0) = Q. Let p be the vector with components given by the coordinates of P. Let:
    [tex] f(t) = |r(t) - p|^2 = (r(t) - p) \cdot (r(t) - p). [/tex]

    What can you say about
    [tex] \frac{d}{dt}f(t) [/tex] at t=0?

    3. The attempt at a solution
    I know that f(t) is at its minimum at t=0 which means f'(t) = 0 at t=0.
    If I let r(t) = <x(t), y(t), z(t)> I get:
    [itex] f'(t) = 2(x(t)-P_x)x'(t) + 2(y(t)-P_y)y'(t) + 2(z(t)-P_z)z'(t) [/itex]

    which gives me:
    [itex] (x(0)-P_x)x'(0) + (y(0)-P_y)y'(0) + (z(0)-P_z)z'(0) = 0 [/itex]

    and since r(0) = Q:
    [itex] (Q_x-P_x)x'(0) + (Q_y-P_y)y'(0) + (Q_z-P_z)z'(0) = 0 [/itex]

    which tells me that [itex] PQ \cdot r'(0) = 0 [/itex], ie PQ is orthogonal to r'(0). This is as far as I have gotten. Any help would be greatly appreciated.
  2. jcsd
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