# What's the point that's the farthest away from the origin on the sphere $(x-1)^2+(y-1)^2+(z-1)^2=1$?

I'm new to 3D-coordinate systems and this is a question I faced in my textbook.

I first tried to imagine the entire thing in 2D: so the same question for the sphere $(x-1)^2+(y-1)^2=1$. It's much simpler to answer this. Keeping in mind that $\sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$, the point is $\left(1 + \frac{1}{\sqrt{2}},1 + \frac{1}{\sqrt{2}}\right)$.

By the same logic, I tried to guess that on the 3D coordinate system, this point would be $\left(1 + \frac{1}{\sqrt{2}},1 + \frac{1}{\sqrt{2}},1 + \frac{1}{\sqrt{2}}\right)$. But this point is not on the sphere.

My intuition tells me that the answer could be $\left(1 + \frac{1}{\sqrt{3}},1 + \frac{1}{\sqrt{3}},1 + \frac{1}{\sqrt{3}}\right)$ and the online 3D grapher I found proved me right. But I don't actually know how to reach this answer.

Any help will be appreciated, thanks.

As Ross says, the point of minimum and maximum distance from origin will lie on line joining centre of sphere with origin.

The centre of sphere is $\langle1,1,1\rangle$ and radius is $1$ unit. So a line passing through origin $O$ and centre $C$ will be $L$, given by:

\begin{align} \vec{L}(r) &= \dfrac{r}{\sqrt{1^2+1^2+1^2}}\langle1,1,1\rangle \\ &= \dfrac{r}{\sqrt{3}} \langle1,1,1\rangle \end{align}

Here, the $r$ is the distance in units. Now, the centre is at $r = \sqrt{3}$. That means the diametrically opposite points along $L$ will be at distance $r_{min}=\sqrt{3} -1$ and $r_{max} =\sqrt{3}+1$ (because radius of sphere is $1$ unit)

Thus $\vec L(r_{min})$ and $\vec L(r_{max})$ represents coordinates, or position vectors of the minimum and maximum distance points.

Draw a line from the origin through the center of the sphere. The point where it hits the sphere on the opposite side from the origin is the farthest point. Your answer is incorrect because the $z$ coordinate of the center is $-1$, not $+1$

• Oh, I made an important typo. Just fixed the title, does this change anything? Commented Oct 10, 2017 at 5:12
• 'important' typo :'D Commented Oct 10, 2017 at 5:12
• Now your answer is correct. You can solve the simultaneous equations for the sphere and the line to verify it. Commented Oct 10, 2017 at 5:15
• @Glycerius then finish by solving $x=y=z$ and $3(x-1)^2=1$ this immediately gives $x=1+\frac 1{\sqrt{3}}$ (sign is + since we search for farthest point).
– zwim
Commented Oct 10, 2017 at 5:37

You want to maximize $x^2+y^2+z^2$ subject to the constraint $(x-1)^2+(y-1)^2+(z-1)^2=1$.

Lagrange mutlipliers are a simple way to do it. Consider the function $$F=x^2+y^2+z^2+\lambda \left((x-1)^2+(y-1)^2+(z-1)^2-1\right)$$ and take derivatives $$\frac{\partial F}{\partial x}=2 \lambda (x-1)+2 x$$ $$\frac{\partial F}{\partial y}=2 \lambda (y-1)+2 y$$ $$\frac{\partial F}{\partial z}=2 \lambda (z-1)+2 z$$ $$\frac{\partial F}{\partial \lambda}=(x-1)^2+(y-1)^2+(z-1)^2-1$$ Eliminate $x,y,z$ for the first partial derivatives (set equal to $0$) and replace in the last. You should get $$\frac{\partial F}{\partial \lambda}=-\frac{\lambda ^2+2 \lambda -2}{(\lambda +1)^2}=0$$ from which $\lambda=-1\pm \sqrt 3$.

I am sure that you can continue from here.

Geometry:

Consider a sphere, centre at origin, radius $R = √3 +1.$

The given sphere has centre at $(1,1,1)$, and radius $r=1$.

The 2 spheres touch each other in one point.

Smaller sphere within larger sphere.

A drawing in 2D may help to visualize.

Furthest point of smaller sphere is $R= √3+1$ away from origin.

http://www.odeion.org/pythagoras/pythag3d.html Max of f(x) s.t (x−1)2+(y−1)2+(z−1)2=1(x−1)2+(y−1)2+(z−1)2=1 ?

max x² + y² + z² Solve for y you get 1 +- sqrt(-x^2 + 2 x - (z - 1)^2) And so on untill you write down y,z as a function of x

Then for an internal max f'=0, f''<0