A vector has both magnitude (size) and direction, unlike a scalar which has only magnitude. You'll meet three notations: a directed segment $\overrightarrow{AB}$ (displacement from $A$ to $B$); an underlined letter in handwriting, $\underline{a}$; and a column vector $\begin{pmatrix}x\\y\end{pmatrix}$, where $x$ is the horizontal shift (positive = right) and $y$ is the vertical shift (positive = up).

Combine component-by-component:

Multiplying by a scalar $k$ scales the length by $|k|$; if $k<0$, the direction also reverses. This is the key to proving two vectors are parallel: if $\mathbf{u}=k\mathbf{v}$ for some number $k$, then $\mathbf{u}$ and $\mathbf{v}$ point along the same line.

The triangle law says a direct journey equals any route via an intermediate point: $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$. The most useful standard result comes from position vectors: if $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ (from origin $O$), then

End minus start — this pattern is worth memorising directly, since it appears constantly.

If $M$ is the midpoint of $AB$, then $\overrightarrow{AM}=\tfrac12\overrightarrow{AB}$. Combined with the triangle law, the position vector of the midpoint is:

A column vector is the hypotenuse of a right-angled triangle, so Pythagoras applies directly:

Example: $\left|\begin{pmatrix}3\\-4\end{pmatrix}\right|=\sqrt{3^2+(-4)^2}=\sqrt{25}=5$.

Worked Example

$OACB$ is a parallelogram, with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. Find $\overrightarrow{OC}$ and $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

1. $\overrightarrow{OC}$: travel via $A$: $\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}$. Since $OACB$ is a parallelogram, $\overrightarrow{AC}=\overrightarrow{OB}=\mathbf{b}$. So $\overrightarrow{OC}=\mathbf{a}+\mathbf{b}$.
2. $\overrightarrow{AB}$: travel via the origin: $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}$.

In a parallelogram $ABCD$, opposite sides are equal in length and direction, so they're represented by the exact same vector: $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{AD}=\overrightarrow{BC}$. This single fact is often enough to find an unknown vector inside a parallelogram without needing to go via the origin at all, and it's also the standard way to prove a quadrilateral is a parallelogram — show both pairs of opposite sides give equal vectors.

If $P$ divides $AB$ in the ratio $2:1$, the segment splits into $2+1=3$ equal parts, so $\overrightarrow{AP}=\tfrac23\overrightarrow{AB}$ and $\overrightarrow{PB}=\tfrac13\overrightarrow{AB}$ — the same reasoning as the midpoint case, just with a different fraction. Always convert the ratio into "parts out of the total" before multiplying.