A position vector fixes a point relative to a chosen origin $O$. The position vector of $A$ is $\overrightarrow{OA}$, written $\mathbf{a}$; the position vector of $B$ is $\overrightarrow{OB} = \mathbf{b}$.

Travel from the start point back to the origin, then out to the end point:

This "route through known points" idea generalises to any path, e.g. $\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC}$, and it's the single most useful technique in this entire topic.

If $P$ lies on $AB$ with $AP:PB = m:n$ (so $m+n$ total parts):

For the midpoint specifically ($m=n=1$): $\overrightarrow{OM} = \dfrac12(\mathbf{a}+\mathbf{b})$.

Two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel exactly when one is a scalar multiple of the other: $\mathbf{u}=k\mathbf{v}$ for some number $k$.

Many shapes give two "base" vectors, usually from a shared vertex, and every other vector in the diagram has to be rebuilt from those two using addition, subtraction and scalar multiples along known routes.

Example: $OABC$ is a parallelogram with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. $X$ is the midpoint of $AB$. Express $\overrightarrow{CX}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

Since $OABC$ is a parallelogram, $\overrightarrow{OB}=\mathbf{a}+\mathbf{c}$ and $AB$ is parallel to $OC$ with $\overrightarrow{AB}=\mathbf{c}$. As $X$ is the midpoint of $AB$:

Then, travelling from $C$ to $X$ via $O$:

Whenever a new letter (like $X$) is introduced partway along an existing side, express its position vector first — every later vector involving that point then becomes a short calculation rather than a fresh diagram-reading exercise.

Two things must be shown, in order:

1. Find two vectors that share a common point (e.g. $\overrightarrow{AB}$ and $\overrightarrow{BC}$), and show one is a scalar multiple of the other — this proves they're parallel.
2. State that because the vectors share the common point $B$, all three points must lie on one straight line.

A scalar-multiple result alone only proves parallelism — the collinearity conclusion needs the "common point" sentence explicitly stated to earn full marks.

In triangle $OAB$, $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. $M$ is the midpoint of $OA$. $N$ lies on $AB$ with $AN:NB = 1:3$. $R$ is the point with $\overrightarrow{OR} = -\dfrac12\mathbf{b}$ (so $R$ lies on line $OB$, on the opposite side of $O$ from $B$). Prove that $M$, $N$ and $R$ are collinear.

Solution

Since $AN:NB=1:3$ (4 parts total), $\overrightarrow{AN} = \dfrac14\overrightarrow{AB} = \dfrac14(\mathbf{b}-\mathbf{a})$, so:

Find $\overrightarrow{MN}$ and $\overrightarrow{MR}$, both starting from the common point $M$:

Since $\overrightarrow{MR} = -2\overrightarrow{MN}$, the two vectors are parallel; since they share the common point $M$, points $M$, $N$ and $R$ lie on a single straight line.