Consider the plane $x+2y+2z=4$, how to find the point on the sphere $x^2+y^2+z^2=1$ that is closest to the plane?
I could find the distance from the plane to the origin using the formula $D=\frac{|1\cdot 0+2\cdot 0+2\cdot 0-4|}{\sqrt{1^2+2^2+2^2}}=\frac43$, and then I can find the distance between the plane and sphere by subtracting the radius of sphere from plane-origin distance:$\frac43-1=\frac13$. But then I am stuck here because I don't know how to convert this distance into a direction vector, so I could subtract it from the plane to find the sphere point. Any help would be appreciated.
Best Answer
The plane unit normal vector is $(1,2,2)$ normalized or $n=(1/3,2/3,2/3)$. Draw a line $l$ through the origin in the direction of $n.$ This line intersects the unit sphere at the point closest to the plane: $(1/3,2/3,2/3).$ (The line $l$ also intersects the unit sphere at $(-1/3,-2/3,-2/3),$ but this is the point on the sphere farthest from the plane)
The reason this is the closest point on the sphere to the plane is that the line $l$ is orthogonal to the tangent plane of the sphere at the point where it intersects the sphere and also orthogonal to the plane.