Triangle ABC is given by overrightarrowOA=vec a…

Triangle $ABC$ is given by $\overrightarrow{OA}=\vec a$, $\overrightarrow{OB}=\vec b$, $\overrightarrow{OC}=\vec c$. Find $\vec r=\overrightarrow{OM}$, where $M$ is an arbitrary point of the perpendicular bisector of $AB$.

Step 1: Equidistance

\[ |\vec r-\vec a|^2=|\vec r-\vec b|^2\iff 2(\vec b-\vec a)\cdot\vec r=|\vec b|^2-|\vec a|^2\iff (\vec b-\vec a)\cdot\left(\vec r-\frac{\vec a+\vec b}{2}\right)=0. \]

So $\overrightarrow{SM}\perp\overrightarrow{AB}$, where $S$ is the midpoint of $AB$. In space this equation alone describes the plane of points equidistant from $A$ and $B$.

Step 2: The bisector in the plane of the triangle

The direction must be perpendicular to $\overrightarrow{AB}$ and lie in the plane $ABC$, whose normal is $\vec N=(\vec b-\vec a)\times(\vec c-\vec a)$: \[\vec t=(\vec b-\vec a)\times\vec N=(\vec b-\vec a)\big[(\vec b-\vec a)\cdot(\vec c-\vec a)\big]-(\vec c-\vec a)\,|\vec b-\vec a|^2.\]

This uses the "BAC–CAB" rule.

\[\vec r=\frac{\vec a+\vec b}{2}+\lambda\Big[(\vec b-\vec a)\big((\vec b-\vec a)\cdot(\vec c-\vec a)\big)-(\vec c-\vec a)|\vec b-\vec a|^2\Big],\quad \lambda\in\mathbb R\]

Because $\vec t\perp\vec b-\vec a$, every such $\vec r$ satisfies

\[ (\vec b-\vec a)\cdot\vec r=\tfrac12\left(|\vec b|^2-|\vec a|^2\right). \]