Hi, I need help with the assignment, Find the point $P$ on the sphere $x^2 + y^2 + z^2 − 8 x − 4 y − 10 z = 4$ which is furthest from the plane Find the point $P$ on the sphere
$x^2 + y^2 + z^2 − 8x − 4y − 10z = 4$
which is furthest from the plane
$6x + 2y − 9z = −23$.
 A: Well, we could start by finding out where the sphere is and how big it is:
$$x^2+y^2+z^2-8x-4y-10z=4\Rightarrow (x-4)^2+(y-2)^2+(z-5)^2=7^2,$$
so its centre $C$ is at $(4,2,5)$ and its radius is $7$.
The equation of the plane is in the form $\textbf{n}\cdot\textbf{r}=k$, where $\textbf{n}$ is a vector normal to the plane, $\textbf{r}$ is the position vector of points on the plane and $k$ is some constant. Thus $\textbf{n}=6\textbf{i}+2\textbf{j}-9\textbf{k}$ is normal to the plane. Normalising this gives $\textbf{n}=\frac{1}{11}(6\textbf{i}+2\textbf{j}-9\textbf{k})$.
Now, thinking geometrically, the vector $\vec{CP}$ should be parallel to $\textbf{n}$ and have length the radius, i.e. 
$$\pmatrix{x-4\\y-2\\z-5}=\pm\frac{7}{11}\pmatrix{6\\2\\-9}.$$
You can go and work out which one of the two solutions given is the point $P$ by using the standard formula for the distance of a point to the plane.
A: Let the point $P \equiv (p, q, r)$ 
The distance of $P$ from $6x+2y−9z=−23$ is given by $\frac{6p+2q−9r+23}{\sqrt{6^2+2^2+9^2}} = \frac{6p+2q−9r+23}{11}$
In other words, we are required to maximize $\frac{6p+2q−9r+23}{11}$ subject to $p^2+q^2+r^2−8p−4q−10r-4= 0$
This is equivalent to the following problem:
Maximize $6p+2q−9r$ subject to $p^2+q^2+r^2−8p−4q−10r-4= 0$
The solution is standard:
Let $f = (6p+2q−9r) + \lambda (p^2+q^2+r^2−8p−4q−10r-4)$
Set $\frac{\partial f}{\partial p} = \frac{\partial f}{\partial q} = \frac{\partial f}{\partial r} = \frac{\partial f}{\partial \lambda} = 0$ and solve.
You'll get $x = \frac{8\lambda -6}{2\lambda}$, $y = \frac{4\lambda -2}{2\lambda}$ and $z = \frac{10\lambda +9}{2\lambda}$ from the first three equations.
Substitute these values into the fourth equation (which is same as the equation of the sphere) and solve for $\lambda$. The solution is $\lambda = \pm \frac{11}{14}$
Corresponding to the two values of $\lambda$, the two possible values of $P$ are $(-\frac{42}{11}, -\frac{14}{11}, \frac{63}{11})$ and $(\frac{42}{11}, \frac{14}{11}, -\frac{63}{11})$ respectively.
