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

Question

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.

Answer 1

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.

Answer 2

WLOG any point on the sphere can be taken as $(\sin t,\cos t\cos u,\cos t\sin u)$

so, the distance will be $$\dfrac{|\sin t+2\cos t(\cos u+\sin u)-4|}{\sqrt{1^2+2^2+2^2}}$$

Now $\cos u+\sin u\le\sqrt2,$ if $\cos t\ge0,$ the equality occurs for $u\equiv\dfrac\pi4\pmod{2\pi}$

$\sin t-2\sqrt2\cos t-4\le\sin t+2\cos t(\cos u+\sin u)-4\le\sin t+2\sqrt2\cos t-4$

$\iff3\sin\left(t-\arccos\dfrac13\right)-4\le\sin t+2\cos t(\cos u+\sin u)-4\le3\sin\left(t+\arccos\dfrac13\right)-4$

Now $\sin\left(t-\arccos\dfrac13\right)\ge-1$

$$\implies\sin t+2\cos t(\cos u+\sin u)-4\ge-7$$

$$\implies|\sin t+2\cos t(\cos u+\sin u)-4|\ge7$$

the equality occurs for $t-\arccos\dfrac13\equiv-\dfrac\pi2\pmod{2\pi}\iff t\equiv-\arcsin\dfrac13$

Answer 3

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.$