Vectors in three-dimensional space: locus of a variable point The fixed point B has position vector $b$ relative to a fixed point $O$. A variable point $M$ has position vector $m$ relative to $O$. Find the locus of M if m $\cdot$ (m – b)=0.
I am told the answer is derived as such:

Can somebody explain the solution to me? I've tried but can't seem to understand...thanks!
 A: If $O$ and $B$ are two antipodal points on a circle in the plane, and $M$ is another point on the circle, then $\angle OMB = 90^\circ$.  In 3-D space, the locus of all such points will be obtained by rotating this circle around the line $OB$, obtaining a sphere.
As an aside, note that you can also solve this problem by completing the square:
\begin{align*}
\vec{m} \cdot \vec{m} - \vec{m} \cdot \vec{b} &= 0 \\
\vec{m} \cdot \vec{m} - \vec{m} \cdot \vec{b} + \frac{1}{4} \vec{b}\cdot \vec{b} &= \frac{1}{4} |\vec{b}|^2 \\
\left( \vec{m} - \frac{1}{2} \vec{b} \right) \cdot \left( \vec{m} - \frac{1}{2} \vec{b} \right) &=  \frac{1}{4} \left| \vec{b} \right|^2 \\
\left| \vec{m} - \frac{1}{2} \vec{b} \right| = \frac{1}{2} \left| \vec{b} \right| 
\end{align*}
In other words, the set of all points that satisfy this equation form a sphere of radius $\frac{1}{2} |\vec{b}|$, centered at the location $\frac{1}{2} \vec{b}$.  This is, of course, equivalent to the answer in the solution you were given.
A: If $\vec{OM}=(x,y,z)$ and $\vec{OB}=(a,b,c)$ then
$$\vec{OM}.\vec{BM}=0\implies$$
$$x(x-a)+y(y-b)+z(z-c)=0 \implies$$
$$(x-\frac{a}{2})^2+(y-\frac{b}{2})^2+(z-\frac{c}{2})^2=\frac{a^2+b^2+c^2}{4}$$
the point M is in the sphere of center $(\frac{a}{2},\frac{b}{2},\frac{c}{2})$ and radius
$R=\frac{\sqrt{a^2+b^2+c^2}}{2}$.
