example of using Lagrange multipliers

One to determine the perpendicular distance of the parallel planes

$$Ax+By+Cz+D=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}Ax+By+Cz+E=\mathrm{\hspace{0.33em}0}$$

is to use the Lagrange multiplier method.  In this case we may to minimise the Euclidean distance of a point $(x,y,z)$  of the former plane to a (fixed) point  $(x_0,y_0,z_0)$  of the latter plane.

Thus we have the equation  $A{x}_0+B{y}_0+C{z}_0+E=\mathrm{\hspace{0.17em}0}$  which we can subtract from the first plane equation, getting

$g:=A(x-x_0)+B(y-y_0)+C(z-x_0)+D-E=\mathrm{\hspace{0.33em}0}.$(1)

This is the (only) constraint equation for minimising the square (http://planetmath.org/SquareOfANumber)

$f:={(x-x_0)}^2+{(y-y_0)}^2+{(z-x_0)}^2$(2)

of the distance of the points.

The polynomial functions $f$ and $g$ satisfy the differentiability requirements.  Accordingly, we can find the minimising point  $(x,y,z)$  by considering the system of equations formed by (1) and

$\{\begin{array}{cc}\frac{\partial f}{\partial x}+\lambda \frac{\partial g}{\partial x}\equiv \mathrm{\hspace{0.33em}2}(x-x_0)+\lambda A=\mathrm{\hspace{0.33em}0},\hfill & \\ \frac{\partial f}{\partial y}+\lambda \frac{\partial g}{\partial y}\equiv \mathrm{\hspace{0.33em}2}(y-y_0)+\lambda B=\mathrm{\hspace{0.33em}0},\hfill & \\ \frac{\partial f}{\partial z}+\lambda \frac{\partial g}{\partial z}\equiv \mathrm{\hspace{0.33em}2}(z-z_0)+\lambda C=\mathrm{\hspace{0.33em}0}.\hfill & \end{array}$(3)

We solve from (3) the differences

$$x-x_0=-\frac{A\lambda }{2},y-y_0=-\frac{B\lambda }{2},z-z_0=-\frac{C\lambda }{2}$$

and set them into (1).  It then yields the value

$$\lambda =\frac{2(D-E)}{A^2+B^2+C^2}$$

of the Lagrange multiplier, which we substitute into the preceding three equations obtaining

$$x-x_0=\frac{A(D-E)}{A^2+B^2+C^2},y-y_0=\frac{B(D-E)}{A^2+B^2+C^2},z-z_0=\frac{C(D-E)}{A^2+B^2+C^2}.$$

These values give the minimal distance when put into the expression of $\sqrt{f}$:

$$d=\sqrt{\frac{{(D-E)}^2(A^2+B^2+C^2)}{{(A^2+B^2+C^2)}^2}}.$$

Hence we have gotten the distance

$$d=\frac{|D-E|}{\sqrt{A^2+B^2+C^2}}.$$