Answer
We show that the minimum distance from the origin to a point on the plane $ax + by + cz = d$ is
$\frac{{\left| d \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
Work Step by Step
We are given the plane $ax + by + cz = d$. Our task is to minimize the distance from the origin $d = \sqrt {{x^2} + {y^2} + {z^2}} $ subject to the constraint $g\left( {x,y,z} \right) = ax + by + cz - d = 0$.
Since finding the minimum distance $d$ is the same as finding the minimum square of the distance ${d^2}$, our task is to minimize $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ subject to the constraint $g\left( {x,y,z} \right) = ax + by + cz - d = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,2y,2z} \right) = \lambda \left( {a,b,c} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $2x = a\lambda $, ${\ \ \ }$ $2y = b\lambda $, ${\ \ \ }$ $2z = c\lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$, $y \ne 0$ and $z \ne 0$.
Thus, $\lambda = \frac{2}{a}x = \frac{2}{b}y = \frac{2}{c}z$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain $y = \frac{b}{a}x$ and $z = \frac{c}{a}x$.
Substituting these in the constraint gives
$ax + by + cz - d = 0$
$ax + \frac{{{b^2}}}{a}x + \frac{{{c^2}}}{a}x - d = 0$
$x\left( {\frac{{{a^2} + {b^2} + {c^2}}}{a}} \right) = d$
So, $x = \frac{{ad}}{{{a^2} + {b^2} + {c^2}}}$.
Using $y = \frac{b}{a}x$ and $z = \frac{c}{a}x$, we obtain the critical point:
$\left( {\frac{{ad}}{{{a^2} + {b^2} + {c^2}}},\frac{{bd}}{{{a^2} + {b^2} + {c^2}}},\frac{{cd}}{{{a^2} + {b^2} + {c^2}}}} \right)$
This critical point corresponds to the minimum of $f$ subject to the constraint $g\left( {x,y,z} \right) = ax + by + cz - d = 0$.
Step 4. Calculate the critical values
We evaluate the minimum value (the minimum square of the distance).
$f\left( {\frac{{ad}}{{{a^2} + {b^2} + {c^2}}},\frac{{bd}}{{{a^2} + {b^2} + {c^2}}},\frac{{cd}}{{{a^2} + {b^2} + {c^2}}}} \right)$
$ = {\left( {\frac{{ad}}{{{a^2} + {b^2} + {c^2}}}} \right)^2} + {\left( {\frac{{bd}}{{{a^2} + {b^2} + {c^2}}}} \right)^2} + {\left( {\frac{{cd}}{{{a^2} + {b^2} + {c^2}}}} \right)^2}$
$ = \frac{{{d^2}\left( {{a^2} + {b^2} + {c^2}} \right)}}{{{{\left( {{a^2} + {b^2} + {c^2}} \right)}^2}}}$
$ = \frac{{{d^2}}}{{{a^2} + {b^2} + {c^2}}}$
Since $f$ is the minimum square of the distance, so the minimum distance from the origin to a point on the plane $ax + by + cz = d$ is
$\frac{{\left| d \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
Notice that there is no maximum value since there are points on the plane that are arbitrarily far from the origin.
