Using Lagrange Multipliers to find the minimum distance of a point to a plane My exercise is as follows: 

Using Lagrange multipliers ﬁnd the distance from the point $(1,2,−1)$
  to the plane given by the equation $x−y + z = 3. $

My thought process:


*

*Langrange Multipliers let you find the maximum and/or minimum of a function given a function as a constraint on your input. For example, if I'm told to find the maximum value of some plane given the constraint $x^2+y^2 = 1$, the only $x$ values I can take are ones on the unit circle.

*It does this by assuming that any intersecting contour lines of the function and constraining function must have the condition $\nabla f = \lambda \nabla g$. Intersecting contour lines are necessary for the constraining function and function to be equal to eachother.

*In this case, we are trying to find the distance from the point $(1,2,-1)$ to the plane of equation $x-y+z=3$. 

*Now, given my intuition, $x-y+z=3$ ought to be $g(x,y,z)$ since in these types of problems some function equal to a constant will be this $g(x,y,z)$ (although I'm not exactly sure why).

*This leads me to ponder what form $f(x,y,z)$ will have such that $F(x,y,z) = f(x,y,z) - \lambda g(x,y,z)$. 

*The minimum distance from a point to a plane should be a straight line, and that line should be perpendicular to the plane. That means it should be the normal vector, or gradient, of that plane. However, I don't know how that helps me. I need a function $f(x,y,z)$ to use.


Here's my lecturer's answer:


*

*We want to minimize $$d = \sqrt {(x-1)^2+(y-2)^2 + (z+1)^2}$$ 

*Equivalent to minimizing $(x-1)^2+(y-2)^2 + (z+1)^2$ with the same constraint.


Thus, it is rendered to
$$F(x,y,z) = (x-1)^2+(y-2)^2 + (z+1)^2 - \lambda (x-y+z-3)$$
Upon solving $\lambda$ apparently didn't need to be considered when minimizing, as it dropped out in row operations and back-substitutions from the simultaneous equations you get from considering the partial derivatives. This leaves me with the following questions:


*

*Why are we working with the distance from the origin to the point rather than the point to the plane?

*Is what is meant by "minimize" is find the smallest output value for the plane given the constraint its inputs must also satisfy $(x-1)^2+(y-2)^2 + (z+1)^2$?

*How did he reach the conclusion $\sqrt{(x-1)^2+(y-2)^2 + (z+1)^2}$ is equivalent to minimizing $(x-1)^2+(y-2)^2 + (z+1)^2$ subject to the same constraint?

*Why is my thinking in normal vectors a bad idea?

*The values that solve the linear equations are the points needed for minimum distance. Why is that?
 A: Vectors is a fine way to solve this.  Find make $(1,2,-1)$ the tail.  Make (1,-1,1) the direction, and find $t$ such that $(1,2,-1) + t\frac {(1,-1,1)}{\|(1,-1,1)\|} = 0$
There is also a distance formula:
What does the old fashioned distance formula say.
$d = \frac {|ax + by + cz - d|}{\sqrt {a^2 + b^2 + c^2}} = \frac {5}{\sqrt {3}}$
Lagrange multiplier are entirely unnecessary in this problem.  But they are useful in more complicated optimization problems.
If you want to use Lagrange multipliers:
$d = \sqrt {(x-1)^2 + (y-2)^2 + (z+1)^2}$ is minimized when $d^2 = (x-1)^2 + (y-2)^2 + (z+1)^2$
We can say this because when $a,b > 0$, and $a>b$ then $a^2 > b^2$
minimizing $d^2$ will minimize $d$
$\frac {\partial F}{\partial x} = 2(x-1) - \lambda = 0\\
\frac {\partial F}{\partial y} = 2(y-2) + \lambda = 0\\
\frac {\partial F}{\partial z} = 2(z+1) - \lambda = 0$
$x = \frac {\lambda + 2}{2}\\
y = \frac {-\lambda  + 4}{2}\\
z = \frac {\lambda - 2}{2}$
plug into the constraint (mutiplying by 2 to kill the deominator):
$2x - 2y + 2z = 3\\
\lambda + 2 + \lambda - 4 + \lambda - 2 = 6\\
\lambda = \frac {10}{3}$
$(x,y,z) = (\frac {8}{3}, \frac {1}{3}, \frac {13}{6})$
$d = \sqrt {\frac {5}{3}^2 + \frac {-5}{3}^2 + \frac {5}{3}^2} = \frac {5\sqrt 3}{3}$
