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.

  • $\begingroup$ Have you learned the Lagrange multiplier? $\endgroup$
    – Frank Lu
    Feb 4, 2017 at 5:24

3 Answers 3


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.


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

  • $\begingroup$ You might want to add a little extra explanation, like sphere x^2+y^2+z^2=1 is the unit sphere and that the you are using polar coordinates (longitude, latitude ) I found it a a really valuable answer, $\endgroup$
    – LOIS 16192
    May 18, 2021 at 6:33

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.

$(2x,2y,2z) = \alpha (1,2,2)$.


$\, 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.$


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