Find the shortest distance from the origin to the surface of $x=yz+10$

Question

we can use the distance formula, but for the sake of simplicity we can use $f(x,y,z)=x^2+y^2+z^2$ and let $\phi(x,y,z)=x-yz=10$. Via Lagrange multiplier, \begin{align}\partial_xf+\partial_x\phi\cdot\lambda=0\\ 2x+\lambda=0\tag{i}\\ \partial_yf+\partial_y\phi\cdot\lambda=0\\ 2y-z\lambda=0\tag{ii}\\ \partial_zf+\partial_z\phi\cdot\lambda=0\\ 2z-\lambda y=0\tag{iii} \end{align}

from equating (ii) and (iii) we find that $$2z=\frac{\lambda^2z}{2}\implies \lambda=\pm 2$$ for $\lambda=2,y=z,$ and $x=-1$, which means $$-1-yz=10$$ $$-yz=11$$ but since $y=z$ $$y^2=-11$$

Therefore $\lambda=-2\implies y=-z,x=1$ \begin{align} -y^2=-9\\ y=3\\ z=-3 \end{align}

Therefore the shortest distance from the origin to the plane is $\sqrt{9+9+1}=\sqrt{19}$

I believe this should be correct, but I have no idea how to verify it, could anyone help me confirm this, or point out where I made a mistake? Thanks!

Answer 1

Here is an alternative approach as a verification. The normal vector of the surface is $(-1,z,y)$. At the point of the shortest distance, the normal vector is parallel to $(x,y,z)$, the direction of the distance line, i.e.

$$\frac x{-1} = \frac yz = \frac zy$$

which leads to the point either at $(-1, y, y) $ or $(1, y, -y)$. Plug them into the surface equation to find that $(-1, y, y) $ is invalid and $(1, y, -y)$ yields $y^2=9$. As a result, the points with the shortest distance are,

$$(1, \pm 3, \mp3)$$

Thus, the shortest distance is $\sqrt{1+9+9}=\sqrt{19}$. Note, the result matches yours, except that it comes from two points on the surface.

Answer 2

We substitute $x = 10 - yz$ into squared distance $d^{2} = x^{2}+y^{2}+z^{2}$ to obtain $ d^{2}=100+20yz+(yz)^{2}+y^{2}+z^{2}$. Rewrite the equation as $d^{2}=19+(y+z)^{2}+(yz+9)^{2}$. This squared distance is minimized if $y+z=0$ and $yz+9=0$, the minimum squared distance is then $19$ and the minimum distance is $\sqrt{19}$.