Finding Point on Plane Closest to Curve

Question

I was given the problem:

Find the point on a plane $4x+5y+z=1$ that is closest to $(23,0,0)$

I am very confused on how to approach this - any ideas?

Answer 1

So you need to minimize $(x-1)^2+(y-0)^2+(z-0)^2$ subject to $6x+7y+1=z$. That should be straight forward.

One easy way is to use the constraint to solve for $z$, substitute and minimize globally over $x,y$.

Answer 2

The normal vector to the plane $$ 6x + 7y + 1 = z$$ is $$N=<6,7,-1>$$

This normal vector is the direction vector of the line perpendicular to the plane.

The parametric equation of the line passing the point $(1,0,0)$ and perpendicular to the plane is $$ x=1+6t, y=7t, z=-t$$

The intersection of this line and the plane is at $$ 6(1+6t)+7(7t)-(-t)=1.$$

Solve for $ t$ and we get $t = \frac {-5}{86}.$

The point of intersection is $P(1+6t,7t,-t)=(\frac {56}{86},\frac {-35}{86}, \frac {5}{86})$ and the distance is $$d=\sqrt {36t^2+49t^2+t^2} = \sqrt {86} t =\frac {5}{\sqrt {86}} \approx0.5391 $$