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"Consider the cone

$$ C := \{ (x,y,z) \in \mathbf{R}^{3} : x^{2}+y^{2} = z^{2}\}$$

and the point $ P:=(1,2,3) $. Find the least diatance between $P$ and any point $Q\in C$ and the coordinates of this point."

Here's what I did. Project everything (the cone and the point $P$) onto the $xy$-plane. $P$ lies on the $y = 2x$ line so if a $Q\in C$ is going to have minimal distance to $P$ it'll lie here as well (right?). I then intersected the cone with the plane $y = 2x$ and locally considered the cone as the line $y = x$, and then found the least distance btw. any point therein to the point $P$ which "locally" had coordinates $(\sqrt{5},3)$.

I found then the least distance to be $\frac{3-\sqrt{5}}{\sqrt{2}}$ and the coordinates of $Q=(x_Q,y_Q,z_Q)$ to be

$$ x_Q = \sqrt{\frac{7+3\sqrt{5}}{10}} $$ $$ y_Q = 2x_Q $$ $$ z_Q = \sqrt{5}x_Q $$

And I come here to ask if this is correct. It might be silly but that square root inside a square root kinda made me think I may have gotten something wrong. Possibly my very first assumption, viz. that $Q$ must lie along the $ y = 2x$ plane was unfounded... I'm going to proofcheck my resolution... The vector between $Q$ and $P$ should be a scalar multiple of the gradient vector of the function of which $C$ is a level surface at $Q$, correct?

wet
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    At least up to the computation of the distance, this seems correct. It may be clearer for you to rotate the system explicitly about the $z$ axis clockwise by $\arctan 2$ (which brings $P$ to $P' = (\sqrt{5}, 0, 3)$, find the intersection there, and then rotate back. This is essentially what you've done, but without using the word "rotation." – Connor Harris Oct 05 '17 at 21:23
  • Looks like you might’ve left out a square root when you computed the least distance. – amd Oct 05 '17 at 21:47
  • Ah, yes. Will fix the OP now. – wet Oct 05 '17 at 22:05
  • It may be worth noting that $\frac12(7+3\sqrt5)=\phi^4,$ where $\phi=\frac12(1+\sqrt5)$ is the golden ratio, so you can rewrite the coordinates without roots inside roots. By the way, $\frac12(3-\sqrt5)=\phi^{-2},$ though that probably doesn't matter much to you. – David K Oct 14 '17 at 20:41
  • No way, quite the contrary – I think it is very cool! How did you notice that? – wet Nov 10 '17 at 01:34

1 Answers1

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The set of normals to the surface of a cone is given by $\nabla f$ where f is the cone. $$f=z-\sqrt{x^2+y^2}$$

$$\nabla f = n = \left ( \frac {-x}{\sqrt{(x^2 + y^2)}}, \frac {-y}{\sqrt{(x^2 + y^2)}},1 \right )$$

Of course we don't know what x & y are, but we can continue anyway and write the line equation from P (1,2,3) to Q (x,y,z) and it would be

$$\left [ \begin{array}{c}x\\y\\z\end{array}\right ] = \left [ \begin{array}{c}1\\2\\3\end{array}\right ] + \lambda \cdot n$$

Now we can split the line equation into three equations, one for each row and then add the cone equation to get 4 equations in 4 unknowns.

The unknowns are x, y, z, and $\lambda$. The equations are $$x = 1 + \lambda n_1$$ $$y=2+\lambda n_2$$ $$z=3+\lambda n_3$$ $$z=\sqrt{x^2+y^2}$$

Their simultaneous solution yields (positives only) $x=(3\sqrt{5}+5)/10\qquad y=(3\sqrt{5}+5)/5$ $z=(\sqrt{5}+3)/2 \qquad \lambda=(\sqrt{5}-3)/2$ These values for x, y, and z are the $Q_x, Q_y, and Q_z$ values and they do match what you found by projecting onto the xy axis. Of course, the "short distance" is just $|P-Q|$.

Narlin
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  • I like this method. Immediately generalises for distance btw. point and any [linear?] surface. – wet Nov 10 '17 at 01:43