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.
$\endgroup$ 13 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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 $t-\arccos\dfrac13\equiv-\dfrac\pi2\pmod{2\pi}\iff t\equiv-\arcsin\dfrac13$
$\endgroup$ 1 $\begingroup$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.$
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