Show that the points are on the sphere In 3-dimension space, the set of all points $\vec{r}$ where $|\vec{r}-\vec{r_1}| = \frac1{2}|\vec{r}-\vec{r_2}|$ happens to be a sphere. 
Show that $\vec{r_3} = \frac2{3}\vec{r_1}+\frac1{3}\vec{r_2}$ and $\vec{r_4} = 2\vec{r_1}-\vec{r_2}$ are both on the sphere.

I'm not sure how to prove this. In fact, I don't think it's true. I get to 
$|\vec{r_3}-\vec{r_1}| = |\frac2{3}\vec{r_1}+\frac1{3}\vec{r_2}-\vec{r_1}| = |-\frac1{3}\vec{r_1}+\frac1{3}\vec{r_2}| = \frac1{3}|-\vec{r_1}+\vec{r_2}|$ which is not equal to $\frac1{2}|\vec{r}-\vec{r_2}|$...
Am I missing something? 
 A: You proved that:
$$|r_3-r_1|=\frac{1}{3}|r_2-r_1|$$
Now $$\frac{1}{2}|r_3-r_2|=\frac{1}{2}|\frac{2}{3}r_1 -\frac{2}{3}r_2|=\frac{1}{3}|r_2-r_1|$$
Thus $|r_3-r_1|=\frac{1}{2}|r_3-r_2|$
Do the same for $r_4$
A: It is true. The calculation for $|\vec{r}_{3} - \vec{r}_{1}|$ is correct:
$$|\vec{r}_{3} - \vec{r}_{1}| = |\frac{2}{3}\vec{r}_{1} + \frac{1}{3} \vec{r}_{2} - \vec{r}_{2}| = |-\frac{1}{3} \vec{r}_{1} + \frac{1}{3} \vec{r}_{2}|.$$
The right side of the equation is
\begin{align}
\frac{1}{2} |\vec{r}_{3} - \vec{r}_{2}| &= \frac{1}{2} |\frac{2}{3}\vec{r}_{1} + \frac{1}{3} \vec{r}_{2} - \vec{r}_{2}| = \\
\frac{1}{2} |\frac{2}{3}\vec{r}_{1} - \frac{2}{3}\vec{r}_{2}| &=
\frac{1}{2} \cdot 2 |\frac{1}{3}\vec{r}_{1} - \frac{1}{3}\vec{r}_{2}| = |\frac{1}{3}\vec{r}_{1} - \frac{1}{3}\vec{r}_{2}|.
\end{align}
Since $|-\frac{1}{3} \vec{r}_{1} + \frac{1}{3} \vec{r}_{2}| = |\frac{1}{3}\vec{r}_{1} - \frac{1}{3}\vec{r}_{2}|$ (their lengths are the same), the given condition is fulfilled. Same applies for $\vec{r}_{4}$:
$$|\vec{r}_{4} - \vec{r}_{1}| = |2\vec{r}_{1} - \vec{r}_{2} - \vec{r}_{1}| = |\vec{r}_{1} - \vec{r}_{2}|$$
and
$$\frac{1}{2} |\vec{r}_{4} - \vec{r}_{2}| = \frac{1}{2} |2\vec{r}_{1} - \vec{r}_{2} - \vec{r}_{2}| = \frac{1}{2} |2\vec{r}_{1} - 2\vec{r}_{2}| = \frac{1}{2} \cdot 2 |\vec{r}_{1} - \vec{r}_{2}|.$$
A: Attempting an interpretation:
Called the circle of Appolonius  in 2 dimensions.
Rotate the circle about its diameter  and get a sphere in 3D.
The problem:
The set of points $\vec r$ with
$\star)$  $|\vec r - \vec r_1| = (1/2)| \vec r - \vec r_2| $,
where $\vec r_1,\vec r_2 $ are fixed points, is a sphere.
If $ \vec r_3,\vec r_4 $ lie on the sphere,  equation $\star)$ is 
satisfied with $\vec r = \vec r_3$ and $\vec r = \vec r_4$ resp.
Check for $\vec r_3 = (2/3)\vec r_1 + (1/3)\vec r_2$.
$\star) LHS $: 
$| \vec r_3 - \vec r_1| = |-(1/3)\vec r_1 +(1/3)\vec r_2|$
$= (1/3)|\vec r_1- \vec r_2|.$
$\star)$ RHS : 
$(1/2) |(2/3)\vec r_1 +(1/3)\vec r_2 - \vec r_2| =$
$(1/2)|(2/3) \vec r_1 - (2/3)\vec r_2| =$
$(1/3)|\vec r_1 - \vec r_2|.$
Check likewise for $ \vec r_4.$
