Position Vectors

Position vectors. Once we fix an origin $O$, every point $P$ in the plane or in space can be represented by a single vector: the vector $\vec{OP}$ is called the position vector of $P$ with respect to $O$. This turns statements about points into statements about vectors, which is what makes the vector method so powerful for geometry.

The fundamental link between vectors and position vectors. For any two points $A,B$ with position vectors $\vec a=\vec{OA}$ and $\vec b=\vec{OB}$: $\vec{AB}=\vec{OB}-\vec{OA}=\vec b-\vec a.$ Proof. By the triangle law $\vec{OA}+\vec{AB}=\vec{OB}$, so $\vec{AB}=\vec{OB}-\vec{OA}$.

Section formula — internal division. Let $P$ divide the segment $AB$ internally in the ratio $m:n$ (i.e. $|AP|:|PB|=m:n$), where $A,B$ have position vectors $\vec a,\vec b$. Then $\vec{OP}=\frac{n\vec a+m\vec b}{n+m}.$ Proof. Write $\vec{OP}=\vec r$. Since $AP:PB=m:n$, we have $n\,|\vec{AP}|=m\,|\vec{PB}|$; and because $\vec{AP}$ and $\vec{PB}$ point the same way (both along $AB$, from $A$ towards $B$), this magnitude relation upgrades to the vector relation $n\,\vec{AP}=m\,\vec{PB}$. Substituting $\vec{AP}=\vec r-\vec a$ and $\vec{PB}=\vec b-\vec r$: $n(\vec r-\vec a)=m(\vec b-\vec r)\ \Longrightarrow\ (n+m)\vec r=n\vec a+m\vec b\ \Longrightarrow\ \vec r=\frac{n\vec a+m\vec b}{n+m}.$

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

Midpoint. Taking $m=n=1$ in the internal formula gives the position vector of the midpoint of $AB$: $\vec{OP}=\frac{\vec a+\vec b}{2}.$

Collinearity of three points. Three distinct points with position vectors $\vec a,\vec b,\vec c$ are collinear if and only if there exist real numbers $x,y,z$, not all zero, such that $x+y+z=0\qquad\text{and}\qquad x\vec a+y\vec b+z\vec c=\vec 0.$ Both conditions are needed — the second alone is too weak (it would also be satisfied by non-collinear points for a suitable, but different, choice of $x,y,z$ that fails the sum-to-zero condition).

Worked idea (dividing a segment given as a combination of two position vectors). If $A,B$ have position vectors $2\vec a+4\vec b$ and $2\vec a-8\vec b$ respectively, the points dividing $AB$ in ratio $1:3$ internally and externally can be found directly by substituting into the section formulas above — the position vectors $\vec a,\vec b$ of the underlying reference points simply carry through the computation unchanged.

Medians of a triangle are concurrent. Let $A,B,C$ have position vectors $\vec a,\vec b,\vec c$, and let $D,E,F$ be the midpoints of $BC,CA,AB$ (so $\vec d=\frac{\vec b+\vec c}2$, etc.). The centroid is the point that divides each median from a vertex to the midpoint of the opposite side internally in the ratio $2:1$. Applying the section formula to median $AD$ with ratio $2:1$: $\vec{OG}=\frac{1\cdot\vec a+2\cdot\vec d}{1+2}=\frac{\vec a+2\left(\frac{\vec b+\vec c}2\right)}3=\frac{\vec a+\vec b+\vec c}{3}.$ Repeating the same computation for median $BE$ and median $CF$ gives exactly the same position vector $\frac{\vec a+\vec b+\vec c}3$ — so all three medians pass through the single point $G$ with $\vec{OG}=\dfrac{\vec a+\vec b+\vec c}3$, proving the medians are concurrent. …