Vectors

A vector describes a displacement: it has both magnitude (length) and direction. Unlike a scalar, direction matters.

Column vector notation: $\mathbf{a}=\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$ means move 3 right and 2 down.

Bold or underlined letter: $\mathbf{a}$ or $\underline{a}$ in print; underline in handwriting.

Directed line segment: $\overrightarrow{AB}$ means the vector from $A$ to $B$.

Magnitude: $|\mathbf{a}|=\left|\left(\genfrac{}{}{0ex}{}{3}{-2}\right)\right|=\sqrt{3^2+(-2{)}^2}=\sqrt{13}$

Translations (G24): translating a shape by vector $\left(\genfrac{}{}{0ex}{}{a}{b}\right)$ moves every point $a$ right and $b$ up (negative values mean left/down).

Adding vectors combines two displacements:

$\left(\genfrac{}{}{0ex}{}{3}{1}\right)+\left(\genfrac{}{}{0ex}{}{-1}{4}\right)=\left(\genfrac{}{}{0ex}{}{2}{5}\right)$

Geometrically: place the second vector's tail at the first vector's head; the result is the vector from the start to the finish.

Subtracting vectors:

$\left(\genfrac{}{}{0ex}{}{5}{2}\right)-\left(\genfrac{}{}{0ex}{}{2}{-1}\right)=\left(\genfrac{}{}{0ex}{}{3}{3}\right)$

Zero vector: $\mathbf{a}+(-\mathbf{a})=\left(\genfrac{}{}{0ex}{}{0}{0}\right)$

Worked example — $\overrightarrow{AB}=\left(\genfrac{}{}{0ex}{}{4}{1}\right)$ and $\overrightarrow{BC}=\left(\genfrac{}{}{0ex}{}{-2}{3}\right)$. Find $\overrightarrow{AC}$.

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\left(\genfrac{}{}{0ex}{}{4}{1}\right)+\left(\genfrac{}{}{0ex}{}{-2}{3}\right)=\left(\genfrac{}{}{0ex}{}{2}{4}\right)$ ✓

Reversing a vector: $\overrightarrow{BA}=-\overrightarrow{AB}=\left(\genfrac{}{}{0ex}{}{-4}{-1}\right)$

Multiplying by a scalar $k$ scales the magnitude by $|k|$ and reverses direction if $k<0$:

$3\left(\genfrac{}{}{0ex}{}{2}{-1}\right)=\left(\genfrac{}{}{0ex}{}{6}{-3}\right)-2\left(\genfrac{}{}{0ex}{}{2}{-1}\right)=\left(\genfrac{}{}{0ex}{}{-4}{2}\right)$

Two vectors are parallel if one is a scalar multiple of the other.

$\left(\genfrac{}{}{0ex}{}{6}{-4}\right)$ is parallel to $\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$ because $\left(\genfrac{}{}{0ex}{}{6}{-4}\right)=2\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$ ✓

Midpoint using vectors: if $M$ is the midpoint of $AB$:

$\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}$

Worked example — $A$ is the origin, $\overrightarrow{AB}=\left(\genfrac{}{}{0ex}{}{8}{4}\right)$. $M$ is the midpoint of $AB$. Find $\overrightarrow{OM}$.

$\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\left(\genfrac{}{}{0ex}{}{0}{0}\right)+\frac{1}{2}\left(\genfrac{}{}{0ex}{}{8}{4}\right)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ✓

A key exam technique: express the vector between two points as a route through known vectors, using addition, subtraction, and scalar multiples.

Golden rule: $\overrightarrow{XY}=\overrightarrow{XO}+\overrightarrow{OY}$ — route from $X$ to $Y$ via $O$. Use $\overrightarrow{XO}=-\overrightarrow{OX}$ to reverse direction.

Worked example — In triangle $OAB$, $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. $M$ is the midpoint of $AB$. Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Route to $M$: go to $A$, then half of $\overrightarrow{AB}$.

$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf{a}+\mathbf{b}$

$\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\mathbf{a}+\frac{1}{2}(-\mathbf{a}+\mathbf{b})=\mathbf{a}-\frac{1}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}=\frac{1}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}$ ✓

Worked example — $P$ divides $AB$ in the ratio $1:2$. Express $\overrightarrow{OP}$.

$\overrightarrow{AP}=\frac{1}{3}\overrightarrow{AB}=\frac{1}{3}(-\mathbf{a}+\mathbf{b})$

$\overrightarrow{OP}=\mathbf{a}+\frac{1}{3}(-\mathbf{a}+\mathbf{b})=\frac{2}{3}\mathbf{a}+\frac{1}{3}\mathbf{b}$ ✓

Vector proofs show geometric properties (midpoints, parallel lines, ratios of lengths) using algebraic manipulation.

Key strategy: express all vectors in terms of two base vectors $\mathbf{a}$ and $\mathbf{b}$. Two line segments are parallel if their vectors are scalar multiples of each other.

Worked example — $O$, $A$, $B$ are points with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. $M$ is the midpoint of $OA$ and $N$ is the midpoint of $OB$. Prove $MN$ is parallel to $AB$ and half its length.

$\overrightarrow{OM}=\frac{1}{2}\mathbf{a}$; $\overrightarrow{ON}=\frac{1}{2}\mathbf{b}$

$\overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\frac{1}{2}\mathbf{b}-\frac{1}{2}\mathbf{a}=\frac{1}{2}(\mathbf{b}-\mathbf{a})$

$\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$

Since $\overrightarrow{MN}=\frac{1}{2}\overrightarrow{AB}$, $MN$ is parallel to $AB$ and half its length. ✓

1. Direction reversal — forgetting the negative sign

$\overrightarrow{BA}=-\overrightarrow{AB}$. Moving from $B$ to $A$ is the opposite direction. Forgetting the negative when reversing a vector is a common error.

2. Column vectors — swapping $x$ and $y$ components

$\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$ means 3 across, 2 down — not 3 up, 2 across. The top number is always the horizontal ($x$) component.

3. Geometric proof — not stating the conclusion explicitly

After showing $\overrightarrow{MN}=k\overrightarrow{AB}$, explicitly state: "Since $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AB}$, $MN$ is parallel to $AB$; since $|k|=\frac{1}{2}$, $MN=\frac{1}{2}AB$." Without this statement, marks are lost.

4. Magnitude — using $x+y$ instead of $\sqrt{x^2+y^2}$

The magnitude of $\left(\genfrac{}{}{0ex}{}{3}{4}\right)$ is $\sqrt{9+16}=5$, not $3+4=7$.

| Mistake | Correction | | ------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------- | --- | ---------- | --- | ---------------------------------------------------------- | | "$\overrightarrow{AC}=\overrightarrow{AB}-\overrightarrow{BC}$" | $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$ (vectors chain, head to tail) | | "$\left(\genfrac{}{}{0ex}{}{2}{-3}\right)$ is parallel to $\left(\genfrac{}{}{0ex}{}{4}{6}\right)$" | $\left(\genfrac{}{}{0ex}{}{4}{6}\right)=-2\left(\genfrac{}{}{0ex}{}{-2}{-3}\right)$, but $\left(\genfrac{}{}{0ex}{}{2}{-3}\right)$ and $\left(\genfrac{}{}{0ex}{}{4}{6}\right)$: $4/2=2$ but $6/(-3)=-2$ — not parallel | | "$|2\mathbf{a}|=2|\mathbf{a}|$" | Correct — magnitude scales by the positive scalar multiple |