Problem source

Let $\mathbf{r}$ be the position vector of a point in three-dimensional space. Describe fully the locus of the point whose position vector is $\mathbf{r}$ in each of the following four cases: 
		\begin{questionparts}
		\item $\left(\mathbf{a-b}\right) \cdot \mathbf{r}=\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2});$
		\item $\left(\mathbf{a-r}\right)\cdot\left(\mathbf{b-r}\right)=0;$
		\item $\left|\mathbf{r-a}\right|^{2}=\frac{1}{2}\left|\mathbf{a-b}\right|^{2};$ 
		\item $\left|\mathbf{r-b}\right|^{2}=\frac{1}{2}\left|\mathbf{a-b}\right|^{2}.$ 
		\end{questionparts}
		Prove algebraically that the equations $\textbf{(i)}$ and $\textbf{(ii)}$ together are equivalent to $\textbf{(iii)}$ and $\textbf{(iv)}$ together.
		Explain carefully the geometrical meaning of this equivalence.

Solution source

\begin{questionparts}
\item $\mathbf{n} \cdot \mathbf{r} = 0$ is the equation of a plane with normal $\mathbf{n}$.

$\mathbf{n} \cdot (\mathbf{r}-\mathbf{a}) = 0$ is the equation of a plane through $\mathbf{a}$ with normal $\mathbf{n}$. 

Our expression is:
\begin{align*}
&& \left(\mathbf{a-b}\right) \cdot \mathbf{r}&=\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) \\
&&&=\frac{1}{2}(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}+\mathbf{b}) \\
\Leftrightarrow &&  \left(\mathbf{a-b}\right) \cdot \left ( \mathbf{r} - \frac12 (\mathbf{a}+\mathbf{b}) \right) &= 0
\end{align*}

So this is a plane through $\frac12 (\mathbf{a}+\mathbf{b})$ perpendicular to $\mathbf{a}-\mathbf{b}$. ie the plane halfway between $\mathbf{a}$ and $\mathbf{b}$ perpendicular to the line between them.

\item \begin{align*}
&& 0 &= \left(\mathbf{a-r}\right)\cdot\left(\mathbf{b-r}\right) \\
&&&= \mathbf{a} \cdot \mathbf{b} - \mathbf{r} \cdot (\mathbf{a}+\mathbf{b}) + \mathbf{r}\cdot\mathbf{r} \\
&&&= \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b})   \right) \cdot \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b})   \right)  - \frac14 \left (\mathbf{a}\cdot\mathbf{a}+2\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{b} \right) +\mathbf{a}\cdot\mathbf{b} \\
&&&= \left | \mathbf{r} - \frac12 \left (\mathbf{a}+\mathbf{b} \right) \right|^2 -  \left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|^2
\end{align*}

Therefore this is a sphere, centre  $\frac12 \left (\mathbf{a}+\mathbf{b} \right)$ radius $\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|$

\item This is a sphere centre $\mathbf{a}$ radius $\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|$

\item This is a sphere centre $\mathbf{b}$ radius $\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|$

\end{questionparts}

Suppose the first two cases are true, then by symmetry it suffices to show that we can prove either of the second cases are true. (Since everything is symmetric in $\mathbf{a}$ and $\mathbf{b}$).

It's useful to note that $\mathbf{r}\cdot \mathbf{r} = \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b}$ from the second condition.

\begin{align*}
\left|\mathbf{r-a}\right|^{2} &= \mathbf{r} \cdot \mathbf{r}-2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\
&= \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b} - 2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\
&= \mathbf{r} \cdot ( \mathbf{b} - \mathbf{a}) + \mathbf{a} \cdot (\mathbf{a}-\mathbf{b}) \\
&= -\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) + |\mathbf{a}|^2- \mathbf{a}\cdot\mathbf{b} \\
&= \frac{1}{2} |\mathbf{a}-\mathbf{b}|^2
\end{align*}

as required.

To show the other direction, consider

Geometrically, these cases are equivalent, because together they both describe a circle of radius $\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|$ in the plane halfway between $\mathbf{a}$ and $\mathbf{b}$