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One way to determine the perpendicular distance of the parallel planes $$Ax+By+Cz+D \;=\; 0 \quad \mbox{and} \quad Ax+By+Cz+E \;=\; 0$$ 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 $Ax_0+By_0+Cz_0+E \,=\, 0$ which we can subtract from the first plane equation, getting
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(1) |
This is the (only) constraint equation for minimising the square
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(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
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(3) |
We solve from (3) the differences $$x-x_0 \;=\; -\frac{A\lambda}{2}, \quad y-y_0 \;=\; -\frac{B\lambda}{2}, \quad z-z_0 \;=\; -\frac{C\lambda}{2}$$ 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 $$x-x_0 \;=\; \frac{A(D-E)}{A^2+B^2+C^2}, \quad y-y_0 \;=\; \frac{B(D-E)}{A^2+B^2+C^2}, \quad z-z_0 \;=\; \frac{C(D-E)}{A^2+B^2+C^2}.$$ 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 formula $$d \;=\; \frac{|D-E|}{\sqrt{A^2+B^2+C^2}}.$$
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