Perpendicular distance from origin to the plane $x+2y+2z=6$

Question

I am trying to find the perpendicular distance from origin to the plane $x+2y+2z=6$. I have an attempt at a solution, but it is incorrect, and I am hoping to find out why.

Consider the function $f(x,y,x)=x^2+y^2+z^2$. We'd like to minimize this function given the contraint $g(x,y,z)=x+2y+2z=6$. To do this, we use the Lagrange multiplier method.

We have $2x=\lambda, 2y=2\lambda, 2z=2\lambda$, from which we get $x=y/2=z/2$. Using this imformation, our constraint equation gives us $y/2+2y+2y=6$, which means that $y=z=4/3$ and $x=2/3$. Plugging this values into $f$, we get $(2/3)^2+(4/3)^2+(4/3)^2=4$.

The answer is supposed to be $2$, so I am not sure where I went wrong. Any help would be much appreciated.

Answer 1

[Turning coffeemath’s comment into an answer since none of the others so far have really addressed your actual question.]

Everything was fine up until the end, when you apparently forgot that the function that you minimized gives you the square of the distance. Take the square root of the value that you’ve computed and all will be well.

Answer 2

Hint: Convert your equation into the Hessian Normalform:

$$\frac{x+2+2z-6}{\pm3}=0$$ Then the distance is given by $$\frac{6}{3}=2$$

Answer 3

You have minimized the wrong function with the Lagrange Multiplier Method.

There is a well-known formula for the distance from a point to a plane.

The formula is $$D= \frac {|ax_0+by_0+cz_0 -d|}{\sqrt {a^2+b^2+c^2}}$$

Where the point is $P= (x_0,y_0,z_0)$ and the equation of the plane is $$ ax+by+cz=d$$

With that formula we find the distance to be $$D=6/3 =2$$