$\mathbf{a} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}$. Find $\mathbf{a} + \mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.
| Multiplier | Effect on the vector |
|---|---|
| $k \gt 1$ | Longer, same direction |
| $0 \lt k \lt 1$ | Shorter, same direction |
| $k = -1$ | Same length, opposite direction |
| $k \lt 0$ | Opposite direction, and scaled by $|k|$ |
$\mathbf{p} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}$. Find $3\mathbf{p} - 2\mathbf{q}$.
Vectors can also be combined geometrically, and this is where most exam marks are.
- Find a route from the start point to the end point along known vectors.
- Write each leg down in order.
- If you travel against an arrow, use the negative of that vector.
- Add the legs together and simplify.
$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. Find $\overrightarrow{AC}$ and $\overrightarrow{OB}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.
In a diagram, $\overrightarrow{PQ} = \mathbf{u}$, $\overrightarrow{QR} = \mathbf{v}$ and $\overrightarrow{RS} = \mathbf{w}$. Express $\overrightarrow{PS}$ and $\overrightarrow{SQ}$ in terms of $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$.
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$. Find $\overrightarrow{OM}$.
$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$, and $P$ lies on $AB$ with $AP : PB = 1 : 3$. Find $\overrightarrow{OP}$.
Sense check: $P$ is closer to $A$, so the coefficient of $\mathbf{a}$ should be the larger one ✓
Vector algebra follows exactly the same rules as ordinary algebra: collect like terms, expand brackets, factorise.
Simplify $3(\mathbf{a} + 2\mathbf{b}) - 2(\mathbf{a} - \mathbf{b})$.
Given $\mathbf{x} + 3\mathbf{a} = 2\mathbf{x} - \mathbf{b}$, find $\mathbf{x}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Adding
Add tops, add bottoms, separately.
Scalar multiple
$k\mathbf{a}$ multiplies both components; always parallel to $\mathbf{a}$.
Triangle law
$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ — middle letters cancel.
Against an arrow
Use the negative: $\overrightarrow{BA} = -\overrightarrow{AB}$.
Parallelogram
Diagonals are $\mathbf{a} + \mathbf{b}$ and $\mathbf{b} - \mathbf{a}$.
Midpoint
$\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ — the average.
Ratio $m:n$
$\overrightarrow{AP} = \dfrac{m}{m+n}\overrightarrow{AB}$.
Algebra
Expand and collect like terms exactly as with numbers.
Route-finding
Any path from start to finish gives the same answer.
$\mathbf{a} = \begin{pmatrix} 6 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$. Find (a) $\mathbf{a} + \mathbf{b}$, (b) $\mathbf{a} - \mathbf{b}$.
▶ Show solution
(a) $\begin{pmatrix} 6 - 2 \\ -1 + 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$
(b) $\begin{pmatrix} 6 - (-2) \\ -1 - 4 \end{pmatrix} = \begin{pmatrix} 8 \\ -5 \end{pmatrix}$
$\mathbf{p} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$. Find (a) $4\mathbf{p}$, (b) $-2\mathbf{p}$, (c) $\tfrac{1}{2}\mathbf{p}$.
▶ Show solution
(a) $\begin{pmatrix} 12 \\ 20 \end{pmatrix}$
(b) $\begin{pmatrix} -6 \\ -10 \end{pmatrix}$
(c) $\begin{pmatrix} 1.5 \\ 2.5 \end{pmatrix}$
$\mathbf{u} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}$. Find $3\mathbf{u} + 2\mathbf{v}$.
▶ Show solution
$3\mathbf{u} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}$ and $2\mathbf{v} = \begin{pmatrix} 10 \\ 2 \end{pmatrix}$
Sum $= \begin{pmatrix} 16 \\ -7 \end{pmatrix}$
Simplify $\overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RS}$.
▶ Show solution
The middle letters cancel in pairs: $P \to Q \to R \to S$.
$= \overrightarrow{PS}$
Simplify $5(\mathbf{a} - 2\mathbf{b}) + 3(2\mathbf{a} + \mathbf{b})$.
▶ Show solution
Expand: $5\mathbf{a} - 10\mathbf{b} + 6\mathbf{a} + 3\mathbf{b}$
Collect: $11\mathbf{a} - 7\mathbf{b}$
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Express $\overrightarrow{AB}$ and $\overrightarrow{BA}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
▶ Show solution
$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$
$\overrightarrow{BA} = \mathbf{a} - \mathbf{b}$
$\overrightarrow{OP} = \mathbf{p}$ and $\overrightarrow{OQ} = \mathbf{q}$. $M$ is the midpoint of $PQ$. Find $\overrightarrow{PM}$ and $\overrightarrow{OM}$.
▶ Show solution
$\overrightarrow{PQ} = \mathbf{q} - \mathbf{p}$
$\overrightarrow{PM} = \tfrac{1}{2}(\mathbf{q} - \mathbf{p})$
$\overrightarrow{OM} = \mathbf{p} + \tfrac{1}{2}(\mathbf{q} - \mathbf{p}) = \tfrac{1}{2}\mathbf{p} + \tfrac{1}{2}\mathbf{q} = \tfrac{1}{2}(\mathbf{p} + \mathbf{q})$
$\mathbf{a} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$. Find the magnitude of $3\mathbf{a}$, and explain the relationship with $|\mathbf{a}|$.
▶ Show solution
$|\mathbf{a}| = \sqrt{16 + 9} = 5$
$3\mathbf{a} = \begin{pmatrix} 12 \\ 9 \end{pmatrix}$, so $|3\mathbf{a}| = \sqrt{144 + 81} = \sqrt{225} = 15$
$15 = 3 \times 5$: multiplying a vector by $k$ multiplies its magnitude by $|k|$.
$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of the diagonal $OB$.
(a) Find $\overrightarrow{OB}$. (b) Find $\overrightarrow{OM}$. (c) Show that $M$ is also the midpoint of $AC$.
▶ Show solution
(a) $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + \mathbf{c}$ (since $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$).
(b) $\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{c})$.
(c) Let $N$ be the midpoint of $AC$. Then $\overrightarrow{ON} = \tfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC}) = \tfrac{1}{2}(\mathbf{a} + \mathbf{c})$.
$\overrightarrow{ON} = \overrightarrow{OM}$, so $N$ and $M$ are the same point.
This proves that the diagonals of a parallelogram bisect each other. ∎
$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. The point $P$ lies on $AB$ with $AP : PB = 2 : 3$. The point $Q$ lies on $OB$ with $OQ : QB = 1 : 4$.
(a) Find $\overrightarrow{OP}$. (b) Find $\overrightarrow{OQ}$. (c) Find $\overrightarrow{QP}$, simplifying your answer.
▶ Show solution
(a) Total parts $= 5$, so $\overrightarrow{AP} = \tfrac{2}{5}\overrightarrow{AB} = \tfrac{2}{5}(\mathbf{b} - \mathbf{a})$.
$\overrightarrow{OP} = \mathbf{a} + \tfrac{2}{5}(\mathbf{b} - \mathbf{a}) = \tfrac{3}{5}\mathbf{a} + \tfrac{2}{5}\mathbf{b}$
(b) Total parts $= 5$, so $\overrightarrow{OQ} = \tfrac{1}{5}\mathbf{b}$.
(c) $\overrightarrow{QP} = \overrightarrow{QO} + \overrightarrow{OP} = -\tfrac{1}{5}\mathbf{b} + \tfrac{3}{5}\mathbf{a} + \tfrac{2}{5}\mathbf{b}$
$= \tfrac{3}{5}\mathbf{a} + \tfrac{1}{5}\mathbf{b}$
Or, factorising: $\overrightarrow{QP} = \tfrac{1}{5}(3\mathbf{a} + \mathbf{b})$