# Proof of orthogonality

1. Oct 11, 2008

### flash

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

$$\frac{d}{dt}f(t)$$ 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:
$f'(t) = 2(x(t)-P_x)x'(t) + 2(y(t)-P_y)y'(t) + 2(z(t)-P_z)z'(t)$

which gives me:
$(x(0)-P_x)x'(0) + (y(0)-P_y)y'(0) + (z(0)-P_z)z'(0) = 0$

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

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