Q.Let $\vec a$ and $\vec b$ be the position vectors of the points $A$ and $B$. Prove that the position vectors of the points which trisect the line segment $AB$ are $\dfrac{2\vec a+\vec b}{3}$ and $\dfrac{\vec a+2\vec b}{3}$.

Fix an origin $O$. For any point $P$, the vector $\vec{OP}$ is the position vector of $P$ with respect to $O$. This single vector encodes the point's entire location, and it converts geometry problems into vector algebra.

The fundamental link. For any two points $A,B$ with position vectors $\vec a=\vec{OA}$, $\vec b=\vec{OB}$: $\vec{AB}=\vec{OB}-\vec{OA}=\vec b-\vec a.$

Section formula (internal division). If $P$ divides segment $AB$ internally in the ratio $m:n$ (i.e. $AP:PB=m:n$), then $\vec{OP}=\frac{n\vec a+m\vec b}{n+m}.$ Idea of the proof: since $\vec{AP}$ and $\vec{PB}$ point the same way and $n|\vec{AP}|=m|\vec{PB}|$, we get $n\,\vec{AP}=m\,\vec{PB}$; writing $\vec{AP}=\vec r-\vec a$ and $\vec{PB}=\vec b-\vec r$ (where $\vec r=\vec{OP}$) and solving gives the formula.

Section formula (external division, without proof). If $P$ divides $AB$ externally in the ratio $m:n$: $\vec{OP}=\frac{m\vec b-n\vec a}{m-n}.$

Midpoint. Setting $m=n=1$ in the internal formula: the midpoint of $AB$ has position vector $\dfrac{\vec a+\vec b}{2}$.

Collinearity test. Three distinct points with position vectors $\vec a,\vec b,\vec c$ are collinear iff there exist real numbers $x,y,z$, not all zero, with $x+y+z=0 \quad\text{and}\quad x\vec a+y\vec b+z\vec c=\vec 0.$

Two classic applications, proved with position vectors:

• Medians of a triangle are concurrent (at the centroid $G$): if $A,B,C$ have position vectors $\vec a,\vec b,\vec c$, the centroid divides each median in ratio $2:1$ from the vertex, and $\vec{OG}=\dfrac{\vec a+\vec b+\vec c}{3}$ — the same point no matter which median you start from.
• A quadrilateral is a parallelogram iff its diagonals bisect each other: $ABCD$ is a parallelogram $\iff \vec a+\vec c=\vec b+\vec d$ (the midpoints of the two diagonals coincide).