Proof about the minimum distance to a plane

Question

How would you use Lagrange multipliers to show that the minimum distance to a plane is given by: $$\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$$ for $P=(x_0,y_0,z_0)$ and the normal vector $n=(a,b,c)$.

Answer 1

Let define $f\colon(x,y,z)\mapsto(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$ and $g\colon (x,y,z)\mapsto ax+by+cz+d$, then: $$\nabla_{(x,y,z)}f=2(x-x_0,y-y_0,z-z_0),\nabla_{(x,y,z)}g=(a,b,c).$$ Therefore, if $(x,y,z)$ is a critical point of $f$ such that $g(x,y,z)=0$, one gets: $$(a,b,c)^{\perp}\subseteq(x-x_0,y-y_0,z-z_0)^{\perp}.$$ In other words, there exists $\lambda\in\mathbb{R}$ such that: $$(x-x_0,y-y_0,z-z_0)=(\lambda a,\lambda b,\lambda c).$$ In particular, one has: $$\begin{align}f(x,y,z)&=\lambda^2(a^2+b^2+c^2),\\a(x-x_0)+b(y-y_0)+c(z-z_0)&=\lambda(a^2+b^2+c^2).\end{align}$$ Hence, since $g(x,y,z)=0$, namely $ax+by+cz=-d$, one gets: $$\lambda=-\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}.$$ This finally leads to: $$f(x,y,z)=\frac{(ax_0+by_0+cz_0+d)^2}{a^2+b^2+c^2}.$$ Whence the result taking the square root.

Remark. This method requires that somehow you have proved that the minimum exists, all I have shown is that $f$ has at most one critical point on $\{g=0\}$ and characterize it.