(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Proof of orthogonality

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