Vectors Revision Notes — 9709 Paper 3

Vectors in 3D: notation, arithmetic and magnitude

Syllabus requirement · §3.7

“

use standard notations for vectors; carry out addition and subtraction of vectors and multiplication of a vector by a scalar, and interpret these operations in geometrical terms; calculate the magnitude of a vector, and use unit vectors, displacement vectors and position vectors. In 2 or 3 dimensions.

”

What a vector is

A scalar is a single number that says how much: a length of $5$ , a temperature of $20^\circ$ . A vector says how much and which way. The easiest way to picture one is as a journey: "walk $3$ units east, then $2$ units north, then climb $4$ units up". The journey has a length and a direction; where you start it from does not change what the journey is.

Every Paper 3 vectors question lives in three dimensions, so we need a way to describe position in space. Take three axes at right angles to each other, all through one point $O$ called the origin:

the $x$ -axis and $y$ -axis make a flat "floor", exactly as on graph paper;
the $z$ -axis points straight up out of that floor.

A point in space then has three coordinates, like $P(3,\,2,\,4)$ : start at $O$ , go $3$ along $x$ , then $2$ along $y$ , then $4$ up in the $z$ direction.

To reach P(3, 2, 4) you travel 3 steps in the i direction, 2 in the j direction and 4 in the k direction. The arrow from O to P is the position vector of P, and it uses exactly the same three numbers.

The notation you must read and write

Three special vectors, each of length $1$ , point along the axes:

\mathbf{i} \text{ along } x, \qquad \mathbf{j} \text{ along } y, \qquad \mathbf{k} \text{ along } z

Because they have length $1$ they are called unit vectors. Any vector is built from them. The journey " $3$ along $x$ , $2$ along $y$ , $4$ up" can be written in any of these ways, and they all mean the same thing:

3\mathbf{i} + 2\mathbf{j} + 4\mathbf{k} \qquad\qquad \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix} \qquad\qquad \overrightarrow{OP}

The first is $\mathbf{i},\mathbf{j},\mathbf{k}$ form.
The second is column form: the top number is the $\mathbf{i}$ part, the middle the $\mathbf{j}$ part, the bottom the $\mathbf{k}$ part.
The third names the vector by its start and end points: $\overrightarrow{OP}$ means "the journey from $O$ to $P$ ".

A single bold letter, such as $\mathbf{a}$ , is also a vector. In handwriting you cannot write bold, so underline it: $\underline{a}$ . A plain $a$ is a number, not a vector.

The numbers $3$ , $2$ and $4$ are the components of the vector. Paper 3 questions switch between the two written forms constantly, sometimes within one question, so practise converting on sight. Column form is usually quicker for arithmetic.

Missing components are zeros. A vector written $4\mathbf{i} - 2\mathbf{j}$ has no $\mathbf{k}$ part, so its column form is $\begin{pmatrix} 4 \\ -2 \\ 0 \end{pmatrix}$ . Forgetting that hidden $0$ is one of the most common slips on the paper — real questions give position vectors like $2\mathbf{i} + 4\mathbf{k}$ (no $\mathbf{j}$ ) or $\mathbf{i} + 2\mathbf{j}$ (no $\mathbf{k}$ ).

Adding, subtracting and scaling

All three operations work component by component — the $\mathbf{i}$ parts together, the $\mathbf{j}$ parts together, the $\mathbf{k}$ parts together:

\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix} \qquad\qquad k\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = \begin{pmatrix} k a_1 \\ k a_2 \\ k a_3 \end{pmatrix}

What addition means. If $\mathbf{u}$ is one journey and $\mathbf{v}$ is another, then $\mathbf{u} + \mathbf{v}$ is "do $\mathbf{u}$ , then do $\mathbf{v}$ " — the single journey that gets you to the same place. Drawn as arrows, the second arrow starts where the first one ends (the triangle law).

What subtraction means. $-\mathbf{v}$ is the journey $\mathbf{v}$ done backwards, so $\mathbf{u} - \mathbf{v}$ means "do $\mathbf{u}$ , then do $\mathbf{v}$ in reverse".

Left: adding is doing one journey after another. Right: to get from A to B you can go back from A to O (that is −a) and then out from O to B (that is b), so AB = b − a.

What scaling means. Multiplying by a number $k$ (a scalar) stretches the arrow by the factor $k$ without turning it. $2\mathbf{v}$ is twice as long as $\mathbf{v}$ and points the same way; $\tfrac12\mathbf{v}$ is half as long; $-\mathbf{v}$ has the same length but points the opposite way.

That gives the most useful fact about scaling:

Two vectors are parallel exactly when one is a number multiple of the other.

To test it, divide matching components. If every component gives the same ratio, the vectors are parallel; if even one ratio differs, they are not. A zero component must be matched by a zero.

v, 2v, ½v and −v all lie along the same line of travel. That is why ‘parallel’ and ‘one is a multiple of the other’ mean the same thing.

Position vector versus displacement vector

A position vector is measured from the origin. $\overrightarrow{OA}$ , usually shortened to $\mathbf{a}$ , says where $A$ is. Its components are the coordinates of $A$ .
A displacement vector joins two points. $\overrightarrow{AB}$ says how to get from $A$ to $B$ .

The right-hand diagram above shows how they are linked. To go from $A$ to $B$ , go from $A$ back to $O$ (the journey $-\mathbf{a}$ ), then from $O$ out to $B$ (the journey $\mathbf{b}$ ):

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

Read it as "end minus start". It is the single most used fact in this topic. Written the wrong way round, $\mathbf{a} - \mathbf{b}$ , it gives the journey from $B$ to $A$ : same length, opposite direction.

Magnitude: the length of a vector

The magnitude (or modulus) of a vector is its length, written $\left\lvert\mathbf{v}\right\rvert$ or $\left\lvert\overrightarrow{AB}\right\rvert$ . The length $AB$ of a line segment means the same thing as $\left\lvert\overrightarrow{AB}\right\rvert$ .

In two dimensions the length of $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2+y^2}$ , straight from Pythagoras. In three dimensions, use Pythagoras twice: once across the floor, then once more going up.

The floor diagonal has length √(x² + y²). It meets the vertical edge z at a right angle, so Pythagoras again gives the full length √(x² + y² + z²).

The computations this whole topic runs on

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

Displacement: end minus start

\left|x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\right| = \sqrt{x^2+y^2+z^2}

Magnitude: Pythagoras in 3D

AB = \left|\mathbf{b} - \mathbf{a}\right|

Distance between two points

\hat{\mathbf{v}} = \frac{1}{\left|\mathbf{v}\right|}\,\mathbf{v}

Unit vector: same direction, length 1

Unit vectors. A unit vector has length exactly $1$ . To get a unit vector in the direction of $\mathbf{v}$ , divide $\mathbf{v}$ by its own length. The direction does not change, because dividing by a positive number is just scaling; the length becomes $\dfrac{\left\lvert\mathbf{v}\right\rvert}{\left\lvert\mathbf{v}\right\rvert} = 1$ . The unit vector is written $\hat{\mathbf{v}}$ ("v hat").

Equal vectors. Two vectors are equal when all three components match. That means they have the same length and the same direction — even if they are drawn in different places. This is what lets us say "opposite sides of a parallelogram are equal vectors" in the next section.

The basic computations on made-up points

The points $A$ and $B$ have position vectors $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 5\mathbf{i} + 3\mathbf{j} - 9\mathbf{k}$ .
(a) Find $2\mathbf{a} - \mathbf{b}$ .
(b) Find $\overrightarrow{AB}$ and the distance $AB$ .
(c) Find a unit vector in the direction of $\overrightarrow{AB}$ .
(d) Show that $\overrightarrow{AB}$ is parallel to $-6\mathbf{i} - 8\mathbf{j} + 24\mathbf{k}$ .

Show full working

1
(a) Write both vectors as columns, remembering every component: $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}, \qquad \mathbf{b} = \begin{pmatrix} 5 \\ 3 \\ -9 \end{pmatrix}$
Column form lines the components up, so each row is its own small sum.
2
Scale $\mathbf{a}$ first — every component doubles: $2\mathbf{a} = \begin{pmatrix} 4 \\ -2 \\ 6 \end{pmatrix}$
3
Subtract row by row: $2\mathbf{a} - \mathbf{b} = \begin{pmatrix} 4 - 5 \\ -2 - 3 \\ 6 - (-9) \end{pmatrix} = \begin{pmatrix} -1 \\ -5 \\ 15 \end{pmatrix} = -\mathbf{i} - 5\mathbf{j} + 15\mathbf{k}$
Subtracting a negative, 6 − (−9), is where signs go wrong. Write the bracket in.
4
(b) End minus start. The journey goes from $A$ to $B$ , so $B$ is the end: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 5 - 2 \\ 3 - (-1) \\ -9 - 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ -12 \end{pmatrix}$
5
The distance $AB$ is the length of that vector. Square each component: $3^2 = 9, \qquad 4^2 = 16, \qquad (-12)^2 = 144$
Squaring removes the minus sign, which is why a negative component never makes a length smaller.
6
Add and square-root: $AB = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
7
(c) Divide the vector by its own length, $13$ : $\hat{\mathbf{u}} = \frac{1}{13}\begin{pmatrix} 3 \\ 4 \\ -12 \end{pmatrix} = \frac{3}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} - \frac{12}{13}\mathbf{k}$
Check: (3/13)² + (4/13)² + (12/13)² = (9 + 16 + 144)/169 = 1, so the length really is 1.
8
(d) Parallel means one is a multiple of the other. Compare the components one pair at a time: $\frac{-6}{3} = -2, \qquad \frac{-8}{4} = -2, \qquad \frac{24}{-12} = -2$
9
All three ratios are the same number, $-2$ , so $-6\mathbf{i} - 8\mathbf{j} + 24\mathbf{k} = -2\,\overrightarrow{AB}$ . The vectors are parallel (the minus sign only means they point opposite ways along the same line).

Answer

(a) $-\mathbf{i} - 5\mathbf{j} + 15\mathbf{k}$ (b) $\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k}$ , $AB = 13$ (c) $\tfrac{1}{13}(3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k})$ (d) it equals $-2\overrightarrow{AB}$

Every later technique in this topic is built from these four moves: combine components, end minus start, Pythagoras in 3D, and ‘is one a multiple of the other?’.

An equal-lengths condition gives an equation

9709/35 M/J 2025 Q10(a)3 marks

With respect to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors given by $\overrightarrow{OA} = 2\mathbf{i} - \mathbf{j} - 6\mathbf{k}, \quad \overrightarrow{OB} = b\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \quad\text{and}\quad \overrightarrow{OC} = -4\mathbf{i} + 5\mathbf{j} - 2\mathbf{k}.$ It is given that $\left\lvert\overrightarrow{AB}\right\rvert = \left\lvert\overrightarrow{BC}\right\rvert$ . Find the value of $b$ .

Show full working

1
Find $\overrightarrow{AB}$ as end minus start, keeping $b$ as a letter: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} b - 2 \\ -2 - (-1) \\ 3 - (-6) \end{pmatrix} = \begin{pmatrix} b - 2 \\ -1 \\ 9 \end{pmatrix}$
2
Find $\overrightarrow{BC}$ the same way: $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} -4 - b \\ 5 - (-2) \\ -2 - 3 \end{pmatrix} = \begin{pmatrix} -4 - b \\ 7 \\ -5 \end{pmatrix}$
3
Equal lengths means equal squared lengths, which avoids square roots: $\left\lvert\overrightarrow{AB}\right\rvert^2 = \left\lvert\overrightarrow{BC}\right\rvert^2$
Both lengths are positive, so squaring both sides cannot create a false solution — and it removes the roots in one move.
4
Write each squared length with Pythagoras: $(b-2)^2 + (-1)^2 + 9^2 = (-4-b)^2 + 7^2 + (-5)^2$
5
Expand the brackets containing $b$ . Note $(-4-b)^2 = (b+4)^2$ : $b^2 - 4b + 4 + 1 + 81 = b^2 + 8b + 16 + 49 + 25$
(−4 − b)² and (b + 4)² are the same because squaring ignores an overall minus sign.
6
Collect the numbers on each side: $b^2 - 4b + 86 = b^2 + 8b + 90$
7
The $b^2$ terms cancel, leaving a linear equation: $-4b + 86 = 8b + 90 \;\Longrightarrow\; -12b = 4 \;\Longrightarrow\; b = -\tfrac13$
Both sides had the same b² term (coefficient 1), which is why the equation became linear and there is only one answer.

Answer

$b = -\tfrac13$

‘Two lengths are equal’ always means: write both displacement vectors, square both magnitudes, set them equal. Never try to handle the square roots directly.

Answer in the form the question uses

Mark schemes are fussy about format in a few specific ways:

A position vector is a vector: write $3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ or a column. Some schemes accept coordinates $(3,\,-1,\,2)$ instead, but some say "Do not accept coordinates" — so give the vector when a vector is asked for.
Never mix the forms: a column with $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ written inside it, or $(3\mathbf{i},\, -\mathbf{j},\, 2\mathbf{k})$ , loses the mark.
When the question says "exact", leave surds such as $\sqrt{14}$ unrounded.

Your turn

Warm-ups on notation and arithmetic, then the unit-vector and equal-lengths styles the papers use.

1
Given $\mathbf{p} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$ and $\mathbf{q} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k}$ , find (a) $\mathbf{p} + \mathbf{q}$ , (b) $3\mathbf{p} - 2\mathbf{q}$ , (c) $\left\lvert\mathbf{q}\right\rvert$ .
Stuck? Show hint
Convert q to a column first. Work row by row.
Show solution
1. 1
  Write $\mathbf{q}$ as a column: $\mathbf{q} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$ .
2. 2
  (a) Add row by row: $\mathbf{p} + \mathbf{q} = \begin{pmatrix} 1 + 3 \\ 0 + (-1) \\ -2 + 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 2 \end{pmatrix}$
3. 3
  (b) Scale each vector first: $3\mathbf{p} = \begin{pmatrix} 3 \\ 0 \\ -6 \end{pmatrix}, \qquad 2\mathbf{q} = \begin{pmatrix} 6 \\ -2 \\ 8 \end{pmatrix}$
4. 4
  Then subtract row by row: $3\mathbf{p} - 2\mathbf{q} = \begin{pmatrix} 3 - 6 \\ 0 - (-2) \\ -6 - 8 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \\ -14 \end{pmatrix}$
5. 5
  (c) Square, add, square-root: $\left\lvert\mathbf{q}\right\rvert = \sqrt{3^2 + (-1)^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$
Answer
(a) $4\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ (b) $-3\mathbf{i} + 2\mathbf{j} - 14\mathbf{k}$ (c) $\sqrt{26}$
29709/33 O/N 2023 Q11(a)2 marks
The line $l$ has equation $\mathbf{r} = \mathbf{i} - 2\mathbf{j} - 3\mathbf{k} + \lambda(-\mathbf{i} + \mathbf{j} + 2\mathbf{k})$ . The points $A$ and $B$ have position vectors $-2\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $3\mathbf{i} - \mathbf{j} + \mathbf{k}$ respectively.

Find a unit vector in the direction of $l$ .
Stuck? Show hint
The direction of a line is the vector multiplied by λ. The fixed point i − 2j − 3k has nothing to do with direction. (Lines are taught properly in “The equation of a line”, further down.)
Show solution
1. 1
  Read off the direction — the vector multiplying $\lambda$ : $\mathbf{d} = -\mathbf{i} + \mathbf{j} + 2\mathbf{k}$
  The part without λ is a point the line passes through; it tells you where the line is, not which way it points.
2. 2
  Find its length: $\left\lvert\mathbf{d}\right\rvert = \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$
3. 3
  Divide the vector by its length: $\hat{\mathbf{d}} = \frac{1}{\sqrt{6}}\left(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\right)$
  The mark scheme also accepts the negative of this, since the line runs both ways.
Answer
$\dfrac{1}{\sqrt6}\left(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\right)$
3
The vectors $\mathbf{u} = 2\mathbf{i} + p\mathbf{j} - 3\mathbf{k}$ and $\mathbf{v} = 6\mathbf{i} - 12\mathbf{j} + q\mathbf{k}$ are parallel. Find $p$ and $q$ .
Stuck? Show hint
Parallel means v = ku for one number k. The i components have no unknowns, so they give k.
Show solution
1. 1
  Parallel means $\mathbf{v} = k\mathbf{u}$ for a single number $k$ : $\begin{pmatrix} 6 \\ -12 \\ q \end{pmatrix} = k\begin{pmatrix} 2 \\ p \\ -3 \end{pmatrix}$
2. 2
  The $\mathbf{i}$ components contain no unknown letter, so use them to find $k$ : $6 = 2k \;\Longrightarrow\; k = 3$
3. 3
  Use $k = 3$ in the $\mathbf{j}$ components: $-12 = 3p \;\Longrightarrow\; p = -4$
4. 4
  Use $k = 3$ in the $\mathbf{k}$ components: $q = 3 \times (-3) = -9$
Answer
$p = -4$ , $q = -9$
4
The points $P$ and $Q$ have position vectors $\mathbf{i} + 2\mathbf{j} + t\mathbf{k}$ and $3\mathbf{i} + 5\mathbf{j} + \mathbf{k}$ , where $t$ is a constant. Given that $PQ = 7$ , find the possible values of $t$ .
Stuck? Show hint
Find PQ in terms of t, then set its squared length equal to 49.
Show solution
1. 1
  End minus start: $\overrightarrow{PQ} = \begin{pmatrix} 3 - 1 \\ 5 - 2 \\ 1 - t \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 1 - t \end{pmatrix}$
2. 2
  Set the squared length equal to $7^2 = 49$ : $2^2 + 3^2 + (1-t)^2 = 49$
3. 3
  Simplify: $4 + 9 + (1-t)^2 = 49$ , so $(1-t)^2 = 36$
4. 4
  Take square roots, remembering both signs: $1 - t = 6 \quad\text{or}\quad 1 - t = -6$
  A squared bracket equal to 36 means the bracket is 6 or −6. Dropping −6 loses one of the two points.
5. 5
  Solve each: $t = -5$ or $t = 7$ .
Answer
$t = -5$ or $t = 7$

Practise vector notation, magnitude and position vectorsReal past-paper questions · Magnitude of a vector; unit, displacement and position vectors

Checking your access…

Can you do all of these?

Convert between column form and i, j, k form — writing missing components as 0
Find AB as b − a, the right way round
Find a length, a distance between two points and a unit vector in 3D
Test whether two vectors are parallel by comparing all three components
Find a midpoint, and a point dividing a segment in a ratio, by walking there
Find the fourth vertex of a parallelogram, rhombus or trapezium, keeping the letters in order round the shape
Write coordinates on a cuboid or pyramid diagram before calculating
Write a line through two points as r = a + t(b − a), always starting ‘r =’
Write the general point of a line in components
Show that a point lies on a line (one parameter value fits all three coordinates)
Use different parameters for two different lines
Check for parallel directions before looking for an intersection
Solve two component equations, then use the third to decide intersecting or skew
Compute a scalar product and use it to test for a right angle
Find the angle between two lines from their directions, and give the acute angle when asked
For angle ABC use BA and BC, both leaving the corner B
Write ‘cos θ = …’ explicitly, and stop there if the cosine is what was asked
Get sin θ exactly from √(1 − cos²θ) and use ½|u||v| sin θ for an area
Double for a parallelogram, or when the angle was taken at a midpoint
Find a foot of a perpendicular with PF · b = 0, then a perpendicular distance as |PF|
Reflect a point with OP′ = OP + 2PF
Turn a perpendicular, intersection or angle condition into an equation in the unknown constant
Square an angle or distance condition, expect two roots, and keep or reject each for a stated reason

Now do the questions

101 real Paper 3 parts from 2021–2025, sorted by difficulty, with mark schemes

Numerical Solution of Equations

Differential Equations

Vectors

Vectors in 3D: notation, arithmetic and magnitude

What a vector is

The notation you must read and write

Adding, subtracting and scaling

Magnitude: the length of a vector

Walking round a shape: midpoints, ratio points, quadrilaterals and solids

The equation of a line: r = a + tb

Can you do all of these?

Vectors in 3D: notation, arithmetic and magnitude

What a vector is

The notation you must read and write

Adding, subtracting and scaling

Magnitude: the length of a vector

The rest of this note

Can you do all of these?