Sec 4 · G3 (Express)

Vectors

Practise vectors with step-by-step help. Free account.

1. Vector notation and magnitude

Symbol	Meaning	Example
$\overrightarrow{AB}$	the vector from $A$ to $B$	$\overrightarrow{CD} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$
$\mathbf{a}$	a vector named by a letter (write $\underline{a}$ by hand)	$\mathbf{a} = \begin{pmatrix} 4 \\ -6 \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix}$	column vector: $x$ across, $y$ up	right and up are positive
$-\mathbf{a}$	same length, opposite direction	$\overrightarrow{BA} = -\overrightarrow{AB}$
$\|\mathbf{a}\|$	magnitude (length)	$\left\|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right\| = 5$

Reading a vector off a grid

\displaystyle \overrightarrow{CD} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

On a grid, CD goes 3 squares across and 2 squares up.

$3$ across and $2$ up

Magnitude

\displaystyle \left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^{2} + y^{2}}

e.g.

\displaystyle \sqrt{4^{2} + (-6)^{2}}

\displaystyle = \sqrt{52}

\displaystyle = 7.21\text{ (3 s.f.)}

An unknown from the magnitude, k > 0

\displaystyle \left|\begin{pmatrix} 12 \\ k \end{pmatrix}\right| = 13

$k^{2} + 12^{2} = 13^{2}$ , so $k^{2} = 25$ and $k = 5$

📖 Worked example

The diagram shows the vector $\overrightarrow{CD}$ on a grid of unit squares. Express $\overrightarrow{CD}$ as a column vector.

Step 1 of 3

Count across from C to D (right is +, left is −)

x\text{-component} = 1

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2. Adding, subtracting and multiples

Add or subtract the components

\displaystyle \mathbf{a} + \mathbf{b}

e.g.

\displaystyle \begin{pmatrix} 4 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ -5 \end{pmatrix}

\displaystyle = \begin{pmatrix} 6 \\ -1 \end{pmatrix}

Triangle law

\displaystyle \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}

AB is a and BC is b; going straight from A to C is a + b.

e.g.

\displaystyle \begin{pmatrix} 2 \\ 6 \end{pmatrix} + \begin{pmatrix} -1 \\ -5 \end{pmatrix}

\displaystyle = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

Scalar multiple: multiply each component

\displaystyle k\mathbf{a}

e.g.

\displaystyle -2\begin{pmatrix} 3 \\ -2 \end{pmatrix}

\displaystyle = \begin{pmatrix} -6 \\ 4 \end{pmatrix}

Parallel vectors: one is a multiple of the other

\displaystyle \mathbf{q} = m\mathbf{p}

$\begin{pmatrix} -6 \\ 4 \end{pmatrix} = -2\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ , so they are parallel

📖 Worked example

$\overrightarrow{AC} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$ . Find $\overrightarrow{BC}$ .

Step 1 of 4

Go from B to C through A

\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}

\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}

📖 Worked example

$\mathbf{p} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} h \\ 4 \end{pmatrix}$ . Given that $\mathbf{p}$ and $\mathbf{q}$ are parallel, find the value of $h$ .

Step 1 of 3

Parallel vectors: how are p and q related?

\mathbf{q} = m\mathbf{p}

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3. Vectors in terms of a and b

Parallelogram OACB with OA = a and OB = b; the diagonal AB is b − a, and P divides AB in the ratio 1 : 2.

Route	Vector
$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}$	$-\mathbf{a} + \mathbf{b}$
$AP : PB = 1 : 2$	$\overrightarrow{AP} = \dfrac{1}{3}\overrightarrow{AB}$
	$\overrightarrow{OP} = \mathbf{a} + \dfrac{1}{3}(-\mathbf{a} + \mathbf{b})$
	$\overrightarrow{OP} = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}$

📖 Worked example

In the diagram, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $P$ lies on $OA$ such that $OP : PA = 1 : 1$ , and $Q$ lies on $OB$ such that $OQ : QB = 1 : 2$ . Express $\overrightarrow{PQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

Step 1 of 3

OP is what fraction of OA? Use the ratio

\overrightarrow{OP} = \dfrac{1}{2}\mathbf{a}

Practise this →

4. Position vectors

Term	Formula	Example: $A(-4, -3)$ , $B(-5, 1)$
position vector of $A$	$\overrightarrow{OA}$	$\begin{pmatrix} -4 \\ -3 \end{pmatrix}$
from $A$ to $B$	$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$	$\begin{pmatrix} -1 \\ 4 \end{pmatrix}$
midpoint $M$ of $AB$	$\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$	$\begin{pmatrix} -4.5 \\ -1 \end{pmatrix}$
length $AB$	$\left\|\overrightarrow{AB}\right\|$	$\sqrt{17}$

📖 Worked example

$\overrightarrow{OA} = \begin{pmatrix} -4 \\ -3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -5 \\ 1 \end{pmatrix}$ . $M$ is the midpoint of $AB$ . Find the position vector of $M$ .

Step 1 of 2

OM is the average of OA and OB

\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})

x: \dfrac{1}{2}(-4 + (-5)) = -4.5

y: \dfrac{1}{2}(-3 + 1) = -1

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5. Vectors in geometry

To show	Show that	Example
parallel	$\overrightarrow{RS} = k\overrightarrow{PQ}$	$6\mathbf{a} - 4\mathbf{b} = 2(3\mathbf{a} - 2\mathbf{b})$
collinear $A$ , $B$ , $C$	$\overrightarrow{AC} = k\overrightarrow{AB}$ (they share $A$ )	$-2\mathbf{a} + 2\mathbf{b} = 2(-\mathbf{a} + \mathbf{b})$

📖 Worked example

$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{OC} = -\mathbf{a} + 2\mathbf{b}$ . Find the value of $k$ such that $\overrightarrow{AC} = k\overrightarrow{AB}$ .

Step 1 of 3

AB = OB − OA

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

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Sec 4 · G3 (Express)

Vectors

Practise vectors with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Vector notation and magnitude

Symbol	Meaning	Example
$\overrightarrow{AB}$	the vector from $A$ to $B$	$\overrightarrow{CD} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$
$\mathbf{a}$	a vector named by a letter (write $\underline{a}$ by hand)	$\mathbf{a} = \begin{pmatrix} 4 \\ -6 \end{pmatrix}$
$\begin{pmatrix} x \\ y \end{pmatrix}$	column vector: $x$ across, $y$ up	right and up are positive
$-\mathbf{a}$	same length, opposite direction	$\overrightarrow{BA} = -\overrightarrow{AB}$
$\|\mathbf{a}\|$	magnitude (length)	$\left\|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right\| = 5$

Reading a vector off a grid

\displaystyle \overrightarrow{CD} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

$3$ across and $2$ up

Magnitude

\displaystyle \left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^{2} + y^{2}}

e.g.

\displaystyle \sqrt{4^{2} + (-6)^{2}}

\displaystyle = \sqrt{52}

\displaystyle = 7.21\text{ (3 s.f.)}

An unknown from the magnitude, k > 0

\displaystyle \left|\begin{pmatrix} 12 \\ k \end{pmatrix}\right| = 13

$k^{2} + 12^{2} = 13^{2}$ , so $k^{2} = 25$ and $k = 5$

📖 Worked example

The diagram shows the vector $\overrightarrow{CD}$ on a grid of unit squares. Express $\overrightarrow{CD}$ as a column vector.

Step 1 of 3

Count across from C to D (right is +, left is −)

x\text{-component} = 1

Practise this →

2. Adding, subtracting and multiples

Add or subtract the components

\displaystyle \mathbf{a} + \mathbf{b}

e.g.

\displaystyle \begin{pmatrix} 4 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ -5 \end{pmatrix}

\displaystyle = \begin{pmatrix} 6 \\ -1 \end{pmatrix}

Triangle law

\displaystyle \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}

e.g.

\displaystyle \begin{pmatrix} 2 \\ 6 \end{pmatrix} + \begin{pmatrix} -1 \\ -5 \end{pmatrix}

\displaystyle = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

Scalar multiple: multiply each component

\displaystyle k\mathbf{a}

e.g.

\displaystyle -2\begin{pmatrix} 3 \\ -2 \end{pmatrix}

\displaystyle = \begin{pmatrix} -6 \\ 4 \end{pmatrix}

Parallel vectors: one is a multiple of the other

\displaystyle \mathbf{q} = m\mathbf{p}

$\begin{pmatrix} -6 \\ 4 \end{pmatrix} = -2\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ , so they are parallel

📖 Worked example

$\overrightarrow{AC} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$ . Find $\overrightarrow{BC}$ .

Step 1 of 4

Go from B to C through A

\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}

\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}

📖 Worked example

$\mathbf{p} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} h \\ 4 \end{pmatrix}$ . Given that $\mathbf{p}$ and $\mathbf{q}$ are parallel, find the value of $h$ .

Step 1 of 3

Parallel vectors: how are p and q related?

\mathbf{q} = m\mathbf{p}

Practise this →

3. Vectors in terms of a and b

Route	Vector
$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}$	$-\mathbf{a} + \mathbf{b}$
$AP : PB = 1 : 2$	$\overrightarrow{AP} = \dfrac{1}{3}\overrightarrow{AB}$
	$\overrightarrow{OP} = \mathbf{a} + \dfrac{1}{3}(-\mathbf{a} + \mathbf{b})$
	$\overrightarrow{OP} = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}$

📖 Worked example

Step 1 of 3

OP is what fraction of OA? Use the ratio

\overrightarrow{OP} = \dfrac{1}{2}\mathbf{a}

Practise this →

4. Position vectors

Term	Formula	Example: $A(-4, -3)$ , $B(-5, 1)$
position vector of $A$	$\overrightarrow{OA}$	$\begin{pmatrix} -4 \\ -3 \end{pmatrix}$
from $A$ to $B$	$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$	$\begin{pmatrix} -1 \\ 4 \end{pmatrix}$
midpoint $M$ of $AB$	$\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$	$\begin{pmatrix} -4.5 \\ -1 \end{pmatrix}$
length $AB$	$\left\|\overrightarrow{AB}\right\|$	$\sqrt{17}$

📖 Worked example

$\overrightarrow{OA} = \begin{pmatrix} -4 \\ -3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -5 \\ 1 \end{pmatrix}$ . $M$ is the midpoint of $AB$ . Find the position vector of $M$ .

Step 1 of 2

OM is the average of OA and OB

\overrightarrow{OM} = \dfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})

x: \dfrac{1}{2}(-4 + (-5)) = -4.5

y: \dfrac{1}{2}(-3 + 1) = -1

Practise this →

5. Vectors in geometry

To show	Show that	Example
parallel	$\overrightarrow{RS} = k\overrightarrow{PQ}$	$6\mathbf{a} - 4\mathbf{b} = 2(3\mathbf{a} - 2\mathbf{b})$
collinear $A$ , $B$ , $C$	$\overrightarrow{AC} = k\overrightarrow{AB}$ (they share $A$ )	$-2\mathbf{a} + 2\mathbf{b} = 2(-\mathbf{a} + \mathbf{b})$

📖 Worked example

Step 1 of 3

AB = OB − OA

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

Practise this →

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