# Use Lagrange Multipliers to show the distance from a point to a plane

I'm trying to use Lagrange multipliers to show that the distance from the point (2,0,-1) to the plane $3x-2y+8z-1=0$ is $\frac{3}{\sqrt{77}}$.

Our professor gave us two hints: We want to minimize a function that describes the distance to (2,0,-1) subject to the constraint $g(x,y,z) = 3x-2y+8z-1=0$, and Compare this method to the equation for measuring the distance from a point to a plane using the normal vector.

I did: Distance = $f(x)= (3x-2)^2 + (2y-0)^2+8z+1)^2$

$g(x) = 3x-2y+8z-1=0$

$\nabla f(x) = ((18x-12),8y,(128z+16)$)

$\nabla g(x) = (3,-2,8)$

Setting them equal to each other and solving $\lambda$:

6x-4 = $6x-4 = \lambda$

-4y = $\lambda$

16z+2 = $\lambda$

Solving these equations simultaneously I get x = $\frac{2}{3}, y = 0, z = \frac{-1}{8}$.

This would mean $\lambda$ is equal to zero, which I know is not possible. It also yields the wrong result when plugged into the distance formula:

$\sqrt{(\frac{2}{3} - 2)^2 + 0^2 + (\frac{-1}{8}+1)^2)} = \sqrt{358}/6$

Can anyone please point out where I am going wrong? Thanks so much!

• For the "distance-squared" function, you just want to write $\ f(x,y,z) = s^2 = (x-2)^2 + (y-0)^2 + (z - [-1])^2 \$ . The constraint function will take care of the rest... That's the beauty of this method! – colormegone Apr 6 '14 at 1:03

The squared distance from $(x,y,z)$ to $(2,0,-1)$ is $(x-2)^2 + (y-0)^2 + (z+1)^2$.