Year 1 Pure · Chapter 15

Vectors

Specification: Edexcel Pure 10 — Vectors · AQA section J

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In this chapter

1. Vectors, components and magnitude
2. Vector arithmetic
3. Position and displacement vectors
4. Ratios and geometric problems
Practice

Key facts — this chapter

A vector has magnitude and direction; in 2D it’s written $a\mathbf{i} + b\mathbf{j}$ or as a column $\begin{pmatrix} a \\ b \end{pmatrix}$ .
Magnitude (length): $\left|\begin{smallmatrix} a \\ b \end{smallmatrix}\right| = \sqrt{a^2 + b^2}$ — Pythagoras on the components.
Add/subtract component by component; multiply by a scalar multiplies each component.
Displacement $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (position of $B$ minus position of $A$ ).
Parallel vectors are scalar multiples of each other: $\mathbf{u} = k\mathbf{v}$ .
A point dividing $AB$ in ratio $m : n$ has position vector $\mathbf{a} + \dfrac{m}{m+n}(\mathbf{b} - \mathbf{a})$ .

A number tells you how much; a vector tells you how much and which way. Vectors package a size and a direction into one object, which makes journeys, forces and displacements clean to calculate with.

1. Vectors, components and magnitude

A 2D vector is described by how far it goes across and up: its components.

Key result

The vector $a\mathbf{i} + b\mathbf{j} = \begin{pmatrix} a \\ b \end{pmatrix}$ goes $a$ across and $b$ up. Its magnitude (length) is

\left|\begin{pmatrix} a \\ b \end{pmatrix}\right| = \sqrt{a^2 + b^2}

— just Pythagoras on the two components.

A vector a from the origin to the point (4, 3), with horizontal component 4, vertical component 3, and magnitude 5 — a = 4i + 3j: 4 across, 3 up, length √(4² + 3²) = 5.

Worked example 15.1 · Magnitude of a vector

Find the magnitude of $\mathbf{v} = 5\mathbf{i} - 12\mathbf{j}$ .

Apply Pythagoras to the components (the signs don’t matter once squared):

|\mathbf{v}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13

2. Vector arithmetic

Vectors add, subtract and scale exactly as you’d hope — one component at a time.

Key result

\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a+c \\ b+d \end{pmatrix}, \qquad k\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}

Geometrically, addition is tip-to-tail: draw the first vector, start the second at its tip, and the resultant runs from the very start to the very end.

Vectors a and b drawn tip to tail, with the resultant a + b from the start of a to the tip of b — Follow a then b; the resultant a + b closes the triangle.

Set the components of a and b. Vector b is drawn from the tip of a, so the resultant a + b runs from the origin straight to the end — tip to tail.

Worked example 15.2 · Combining vectors

Given $\mathbf{a} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$ , find $2\mathbf{a} - \mathbf{b}$ .

Scale $\mathbf{a}$ by $2$ :
$2\mathbf{a} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$
Subtract $\mathbf{b}$ component by component:
$2\mathbf{a} - \mathbf{b} = \begin{pmatrix} 6 \\ 2 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 0 \end{pmatrix}$

3. Position and displacement vectors

The position vector of a point $A$ is $\overrightarrow{OA} = \mathbf{a}$ , the vector from the origin to $A$ . The vector between two points comes from subtracting their positions.

Key result

\overrightarrow{AB} = \mathbf{b} - \mathbf{a} \qquad (\text{position of } B \text{ minus position of } A)

The distance $AB$ is then $\left|\overrightarrow{AB}\right|$ .

Worked example 15.3 · Displacement and distance

Points $A$ and $B$ have position vectors $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ . Find $\overrightarrow{AB}$ and the distance $AB$ .

Displacement is “end minus start”:
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$
Distance is its magnitude:
$AB = \sqrt{3^2 + 4^2} = 5$

Common mistake

$\overrightarrow{AB}$ is $\mathbf{b} - \mathbf{a}$ , not $\mathbf{a} - \mathbf{b}$ — end minus start. Getting it backwards gives $\overrightarrow{BA}$ , the same length but the opposite direction.

4. Ratios and geometric problems

Vectors shine at “prove these points are collinear” and “find the point dividing this line in a given ratio” — the kind of geometry that’s fiddly with coordinates alone.

Parallel, and dividing a line

Two vectors are parallel exactly when one is a scalar multiple of the other: $\mathbf{u} = k\mathbf{v}$ . (If they also share a point, that makes the points collinear.)
The point $P$ dividing $AB$ in the ratio $m : n$ has position vector
$\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a}).$
(With $m = n$ this is the midpoint, $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .)

Points A and B with a point P on segment AB dividing it in the ratio 2 to 1, with position vectors from the origin O — P divides AB in the ratio 2 : 1, two-thirds of the way from A to B.

Worked example 15.4 · Dividing a line in a ratio

Points $A$ and $B$ have position vectors $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 8 \\ 7 \end{pmatrix}$ . Find the position vector of the point $P$ that divides $AB$ in the ratio $2 : 1$ .

$P$ is $\dfrac{2}{2+1} = \dfrac{2}{3}$ of the way from $A$ to $B$ :
$\overrightarrow{OP} = \mathbf{a} + \tfrac{2}{3}(\mathbf{b} - \mathbf{a})$
Compute $\mathbf{b} - \mathbf{a} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}$ , so $\tfrac{2}{3}(\mathbf{b} - \mathbf{a}) = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$ :
$\overrightarrow{OP} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}$

Worked example 15.5 · Showing points are collinear

Show that $A(1, 2)$ , $B(3, 5)$ and $C(7, 11)$ are collinear.

Find two displacement vectors from $A$ :
$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}, \qquad \overrightarrow{AC} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$
$\overrightarrow{AC} = 3\overrightarrow{AB}$ , so the two vectors are parallel — and since they share the point $A$ , the three points lie on one straight line. ∎

Common mistake

Parallel vectors alone don’t prove collinearity: you also need a shared point. $\overrightarrow{AB} \parallel \overrightarrow{CD}$ just means two separate lines are parallel; it’s $\overrightarrow{AB} \parallel \overrightarrow{AC}$ (same starting point $A$ ) that forces $A$ , $B$ , $C$ onto one line.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Find the magnitude of each vector:

$\begin{pmatrix} 6 \\ 8 \end{pmatrix}$
$3\mathbf{i} - 4\mathbf{j}$

Show solution

(a) $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ .

(b) $\sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$ .

2 [3 marks]

Given $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ , find:

$\mathbf{a} + \mathbf{b}$
$3\mathbf{a}$
$\mathbf{a} - \mathbf{b}$

Show solution

(a) $\begin{pmatrix} 2+3 \\ -1+4 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$ .

(b) $\begin{pmatrix} 6 \\ -3 \end{pmatrix}$ .

(c) $\begin{pmatrix} 2-3 \\ -1-4 \end{pmatrix} = \begin{pmatrix} -1 \\ -5 \end{pmatrix}$ .

3 [3 marks]

Points $A$ and $B$ have position vectors $\mathbf{a} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 7 \\ 11 \end{pmatrix}$ . Find $\overrightarrow{AB}$ and the distance $AB$ .

Show solution

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$ , and $AB = \sqrt{6^2 + 8^2} = 10$ .

4 [2 marks]

Show that $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 8 \\ 12 \end{pmatrix}$ are parallel.

Show solution

$\begin{pmatrix} 8 \\ 12 \end{pmatrix} = 4\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ — one is a scalar multiple of the other, so they are parallel.

Standard

5 [3 marks]

Find a unit vector in the direction of $\mathbf{v} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}$ .

Show solution

The magnitude is $|\mathbf{v}| = \sqrt{5^2 + (-12)^2} = 13$ . A unit vector divides by the magnitude:

\hat{\mathbf{v}} = \frac{1}{13}\begin{pmatrix} 5 \\ -12 \end{pmatrix} = \begin{pmatrix} 5/13 \\ -12/13 \end{pmatrix}

6 [3 marks]

Given $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ , find $2\mathbf{a} + 3\mathbf{b}$ .

Show solution

2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 2 \\ 4 \end{pmatrix} + \begin{pmatrix} 12 \\ -3 \end{pmatrix} = \begin{pmatrix} 14 \\ 1 \end{pmatrix}

7 [4 marks]

The vector $\begin{pmatrix} 6 \\ p \end{pmatrix}$ is parallel to $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ . Find the value of $p$ .

Show solution

For parallel vectors, $\begin{pmatrix} 6 \\ p \end{pmatrix} = k\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ . The first components give $6 = 2k$ , so $k = 3$ . Then $p = 5k = 15$ .

8 [4 marks]

Points $A$ and $B$ have position vectors $\mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 9 \\ 8 \end{pmatrix}$ . Find the position vector of the midpoint $M$ of $AB$ .

Show solution

\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \tfrac{1}{2}\begin{pmatrix} 12 \\ 10 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}

Exam style

9 [5 marks]

Points $A$ , $B$ and $C$ have position vectors $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ , $\begin{pmatrix} 4 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 8 \\ 10 \end{pmatrix}$ .

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . [2]
Hence show that $A$ , $B$ and $C$ are collinear, and state the ratio $AB : BC$ . [3]

Show solution

(a) $\overrightarrow{AB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ , $\overrightarrow{AC} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$ .

(b) $\overrightarrow{AC} = 3\overrightarrow{AB}$ , so the vectors are parallel and share the point $A$ — hence $A$ , $B$ , $C$ are collinear. Since $\overrightarrow{AC} = 3\overrightarrow{AB}$ , $B$ is one-third of the way along, so $AB : BC = 1 : 2$ .

10 [5 marks]

$\overrightarrow{OA} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$ . The point $P$ divides $AB$ in the ratio $1 : 2$ .

Find $\overrightarrow{OP}$ . [3]
Find the exact distance $OP$ . [2]

Show solution

(a) $P$ is $\tfrac{1}{3}$ of the way from $A$ to $B$ :

\overrightarrow{OP} = \mathbf{a} + \tfrac{1}{3}(\mathbf{b} - \mathbf{a}) = \begin{pmatrix} 1 \\ 4 \end{pmatrix} + \tfrac{1}{3}\begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}

(b) $OP = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ .

11 [5 marks]

$\overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}$ . Point $C$ is such that $OACB$ is a parallelogram (with $OA$ and $OB$ as adjacent sides). Find the position vector of $C$ .

Show solution

In parallelogram $OACB$ , $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB}$ (the diagonal is the sum of the two sides):

\overrightarrow{OC} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}

12 [5 marks]

The vector $\mathbf{c} = \begin{pmatrix} k \\ k + 1 \end{pmatrix}$ has magnitude $5$ . Find the possible values of $k$ .

Show solution

Set the magnitude squared equal to $25$ :

k^2 + (k + 1)^2 = 25

Expand and tidy: $k^2 + k^2 + 2k + 1 = 25$ , so $2k^2 + 2k - 24 = 0$ , i.e. $k^2 + k - 12 = 0$ .

Factorise: $(k + 4)(k - 3) = 0$ , so $k = 3$ or $k = -4$ .

(Check: $k = 3$ gives $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $k = -4$ gives $\begin{pmatrix} -4 \\ -3 \end{pmatrix}$ — both of magnitude $5$ .)