Determinate the furthest and nearest points on an ellipsoid from a plane

Question

I'm a bit confused in the methods that could help to solve this. The problem is the following:

In the ellipsoid: $$4x^2+y^2+4z^2-16x-6y-8z+25=0$$ find the furthest point and the nearest point of the plane: $$2x+2y+z=0$$

I think that using the method of Lagrange multipliers wil solve this, but I don't even know how to start. Can I give a suggestion for this? Please, and thanks.

Answer 1

The vector normal to the plane $2x+2y+z=0$ is (2,2,1). The normal vector to the ellipsoid at point $(x,y,z)$ is $(z_x',z_y',-1)$, or explicitly,

$$\left(-\frac{x-2}{z-1}, -\frac{y-3}{4z-4},-1\right)$$

For the nearest and furthest points from the plane, the normal vectors at these two points are parallel to the normal vector of the plane, i.e.

$$-\frac{x-2}{z-1}:-\frac{y-3}{4z-4}:-1 = 2:2:1$$

which leads to,

$$x=2z,\>\>\>\>\>y=8z-5\tag{1}$$

Substitute (1) into the equation of the ellipsoid $4x^2+y^2+4z^2-16x-6y-8z+25=0$ to get

$$21z^2-42z+20=0$$

which yields the coordinates $z=1\pm\frac 1{\sqrt{21}}$. Plug $z$ back into (1) to obtained the nearest and furthest point at

$$\left(2\pm\frac 2{\sqrt{21}},3\pm\frac 8{\sqrt{21}},1\pm\frac 1{\sqrt{21}}\right)$$

Answer 2

Hints:

1. The distance from $\bf p$ to the plane ${\bf a \cdot \bf x} = 0$ is $|{\bf a \cdot \bf p}|/|{\bf a}|$, where $|{\bf a}|$ is the length of $\bf a$.
2. A point on the intersection of the plane with the ellipsoid (if such exists) will minimize the distance.
3. The maximum distance is obtained by minimizing or maximizing ${\bf a \cdot \bf p}$ on the ellipsoid. Thus, using a Lagrange multiplier, you take $F(x,y,z,\lambda) = \lambda (4 x^2 + \ldots + 25) + 2 x + 2 y + z$ and find its critical points.