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:
Prove algebraically that the equations \((i)\) and \((ii)\) together are equivalent to \((iii)\) and \((iv)\) together.
Explain carefully the geometrical meaning of this equivalence.
Solution:
\(\mathbf{n} \cdot \mathbf{r} = 0\) 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.
This is a sphere centre \(\mathbf{a}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)
This is a sphere centre \(\mathbf{b}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)
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 \(\mathbf{b}\)