Distance from point to cone "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?
 A: 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 $|P-Q|$. 
