Position Vectors

The position vector of a point $A$ is $\overrightarrow{OA}$, the vector from the origin $O$ to $A$. It is conventionally written as the lower-case bold letter matching the point, so point $A$ has position vector $\mathbf{a}$.

In components the position vector simply repeats the coordinates: the point $\left(3,-2\right)$ has position vector $3\mathbf{i}-2\mathbf{j}$. That is why position vectors let you do coordinate geometry with vector algebra — the two notations carry identical information.

To travel from $A$ to $B$, go backwards along $\mathbf{a}$ to the origin, then forwards along $\mathbf{b}$. So $\overrightarrow{AB}=-\mathbf{a}+\mathbf{b}$, usually written the other way round.

It is worth reading the subtraction order carefully: the vector points $\text{to}$ the second letter, so the second letter’s position vector comes first. $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$, and $\overrightarrow{BA}=\mathbf{a}-\mathbf{b}$, which is its negative.

The midpoint of $AB$ has position vector $\frac{1}{2}\left(\mathbf{a}+\mathbf{b}\right)$ — the average of the two, exactly as with coordinates.

For a general ratio, think of it as a journey: start at $A$ and travel a fraction of the way along $\overrightarrow{AB}$. A point dividing $AB$ in the ratio $m:n$ is $\frac{m}{m+n}$ of the way along, so its position vector is $\mathbf{a}+\frac{m}{m+n}\left(\mathbf{b}-\mathbf{a}\right)$.

Deriving it that way is more reliable than memorising a formula, because the "start plus a fraction of the journey" picture makes the direction obvious.

Three points are $\text{collinear}$ when they lie on one straight line. In vector terms, two displacement vectors between them must be parallel — and because they share a point, parallel is enough to force them onto the same line.

So to show $A$, $B$ and $C$ are collinear, compute $\overrightarrow{AB}$ and $\overrightarrow{BC}$ and show one is a scalar multiple of the other. Stating the shared point is part of the argument: parallel alone would only give two parallel lines.