How to find the point on the sphere that is closest to a plane? 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.

A little different:
Plane: $f(x,y,z) = x + 2y + 2z - 4 = 0 $; Circle $ g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 $.
Problem:
Point on a sphere with minimum distance to the plane.
Normal vector to the plane, $n_p$:
$\nabla f = (\frac {\partial f}{\partial x},\frac {\partial f}{\partial y},\frac {\partial f}{\partial z})$.
We get $n_p = (1,2,2)$.
Normal vector to the circle, $ n_c$:
$\nabla g = (\frac {\partial g}{\partial x},\frac {\partial g}{\partial y},\frac {\partial g}{\partial z})$.
We get $n_c = (2x,2y,2z)$.
At the desired point on the circle the two normals are parallel or anti parallel.
$(2x,2y,2z) = \alpha (1,2,2)$.
Hence:
$\, 2x = \alpha, 2y = 2\alpha , 2z = 2\alpha $.
Combining the above with the equation of the circle:
$ \alpha ^2 + ( 2\alpha)^2 + (2\alpha)^2 = 4$,
$ \alpha ^2 + 4 \alpha ^2 + 4 \alpha ^2 = 4$,
$ 9 \alpha ^2 = 4$,
$ \alpha_1 = 2/3$ and $\alpha_2 = - 2/3$.
The closest point:
1) $P_1 (1/3,2/3,2/3)$ for $\alpha_1 = 2/3$, 
the farthest:
2) $P_2 (-1/3,-2/3,-2/3)$ for $ \alpha_2 = - 2/3.$
