Let's do this with Lagrange Multipliers.
We ought to use the distance formula if we want to minimize the distance.
Right?.
Well, we're gonna dodge the radicals and minimize the square of the
distance to (1,2,3).
You want to minimize \(\displaystyle \L\\f(x,y,z)=(x-1)^{2}+(y-2)^{2}+(z-3)^{2}\)
subject to the constraint:
\(\displaystyle \L\\g(x,y,z)\Rightarrow{z^{2}-x^{2}-y^{2}=0}\)
Partial derivatives:
\(\displaystyle \L\\2(x-1)+2(y-2)+2(z-3)={\lambda}(-2x-2y+2z)\)
By equating like coefficients, this gives us:
\(\displaystyle \L\\2(x-1)=-2x{\lambda}\\2(y-2)=-2y{\lambda}\\2(z-3)=2z{\lambda}\)
Therefore,
\(\displaystyle \L\\\frac{1-x}{x}={\lambda}...[1]\\\frac{2-y}{y}={\lambda}...[2]\\
\frac{z-3}{z}={\lambda}....[3]\)
Equate [1] and [2]:
\(\displaystyle \L\\\frac{1-x}{x}=\frac{2-y}{y}\)
\(\displaystyle \L\\y=2x.........[4]\)
Equate [1] and [3]:
\(\displaystyle \L\\\frac{1-x}{x}=\frac{z-3}{z}\)
\(\displaystyle \L\\z=\frac{3x}{2x-1}.....[5]\)
Sub into g(x,y,z):
\(\displaystyle \L\\(\frac{3x}{2x-1})^{2}-x^{2}-(2x)^{2}=0\)
Solvng for x:
\(\displaystyle \L\\x=(\frac{3\sqrt{5}+5}{10}\,\ \frac{-(3\sqrt{5}-5)}{10}\,\ 0)\)
Subbing into [4] and [5]:
\(\displaystyle \L\\y=(\frac{3\sqrt{5}+5}{5}\,\ \frac{-(3\sqrt{5}-5)}{5}\,\ 0)\)
\(\displaystyle \L\\z=(\frac{\sqrt{5}}{2}+\frac{3}{2}\,\ \frac{3}{2}-\frac{\sqrt{5}}{2}\,\ 0)\)
Now, we still need to check and see which ones are the closest.
By subbing these into f(x,y,z) we find that:
Drum roll, please..............................
\(\displaystyle \H\\x=\frac{3\sqrt{5}+5}{10}=\approx{1.17}\\\\y=\frac{3\sqrt{5}+5}{5}\approx{2.34}\\\\z=\frac{\sqrt{5}+3}{2}\approx{2.62}\)
is the point on the cone closest to (1,2,3).
