# Thread: find distance from plane to point

1. ## find distance from plane to point

the plane given is

$\displaystyle x+2y+3z-10=0$

distance formula is $\displaystyle \sqrt{x^2+y^2+z^2}$

I know that the distance squared will have the same minimizers so I'll use that

I need to solve for z to get it into a two variable problem

$\displaystyle z= \frac{-1-2y+10}{3}$

now,

my squared distance is

$\displaystyle x^2+y^2+\left(\frac{-x-2y-10}{3}\right)^2$

now take gradient, solve for minimizers then sub-in and find distance

correct?

2. ## Re: find distance from plane to point

Yes, you can do that, If the given point is the origin (0, 0, 0). You don't say that. If the given point is $\displaystyle (x_0, y_0, z_0)$ then the function to be minimized is $\displaystyle (x- x_0)^2+ (y- y_0)^2+ \left(\frac{-x- 2y-10}{3}- z_0\right)^2$

Personally, I would NOT do it that way. Geometrically, the shortest distance from a point to a plane is along a line through the given point perpendicular to the given line. That means it must lie in the plane containing the given point, perpendicular to the given line. Here, the given plane is x+ 2y+ 3z- 10= 0 so a perpendicular vector is given by <1, 2, 3> and the line through $\displaystyle (x_0, y_0, z_0)$ parallel to that vector, and so perpendicular to the plane, is $\displaystyle x= t+ x_0$, $\displaystyle y= 2t+ y_0$, $\displaystyle z= 3t+ z_0$. Determine the point where that line crosses the given plane and find the distance from that point to $\displaystyle (x_0, y_0, z_0)$.

If the given point really is (0, 0, 0), the perpendicular line is given by x= t, y= 2t, z= 3t. Determine where that crosses the given plane by replacing x, y, and z in the equation of the plane by those: x+ 2y+ 3z- 10= t+ 2(2t)+ 3(3t)- 10= 14t- 10= 0. Solve that for t.

3. ## Re: find distance from plane to point

sorry, yeah I meant the origin