Notes · MathematicsUK · A-Levels

Vectors

Vectors describe quantities with both magnitude and direction in two and three dimensions. The A-Level covers vector notation, magnitude and direction, addition, subtraction and scalar multiplication, unit and position vectors, and the use of vectors to solve geometric and kinematic problems. (The scalar product and vector equations of lines belong to Further Mathematics.)

4 sections~9 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

T·111111 / 20

Exam profile

AO1 · Carry out vector arithmetic and find magnitudes in 2D and 3DAO2 · Prove geometric relationships such as collinearity and ratios using vectorsAO3 · Model positions and displacements with vectors

Operators:findshow thatprovedetermineexpress

basic level

AS-Level covers 2D vectors, magnitude and direction, addition, scalar multiples and position vectors.

higher level

The full A-Level extends all of this to three dimensions and to geometric proofs (collinearity, ratios, midpoints).

Depth

Text

Contents · 4 sections

Vectors

Vectors, notation and magnitude#

FoundationDfE GCE Mathematics J1AQA 7357 J

A vector and its components

Fig. 1

\mathbf{a} = 4\mathbf{i} + 3\mathbf{j}

has magnitude

\sqrt{4^2 + 3^2} = 5

A vector is a quantity with both magnitude and direction, in contrast to a scalar, which has magnitude only. Displacement, velocity, acceleration and force are vectors; distance, speed and mass are scalars. Vectors are written in bold (

\mathbf{a}

) or underlined, and a displacement from

A

B

\overrightarrow{AB}

. This distinction between what has direction and what does not runs through all of mechanics.

In component form a two-dimensional vector is written either as a column

\begin{pmatrix} a \\ b \end{pmatrix}

or in terms of the unit vectors

\mathbf{i}

(one unit in the

x

-direction) and

\mathbf{j}

(one unit in the

y

-direction) as

a\mathbf{i} + b\mathbf{j}

. The two notations are interchangeable, and three-dimensional vectors simply add a

z

-component with the unit vector

\mathbf{k}

The magnitude (or modulus) of a vector is its length, found by Pythagoras:

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

in two dimensions, and

\sqrt{a^2 + b^2 + c^2}

in three. This is the distance the displacement covers, or the size of the velocity or force. The direction of a 2D vector can be given as the angle it makes with the positive

x

-axis, found using

\tan\theta = \dfrac{b}{a}

Two vectors are equal only if they have the same magnitude and the same direction — position in the plane is irrelevant, so a vector can be translated freely. A vector's negative

-\mathbf{a}

has the same magnitude but the opposite direction. These conventions let the same displacement be represented wherever it is convenient, which is what makes vector geometry so flexible.

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

Magnitude of a vector

The length is found by Pythagoras from the components.

Worked example

Magnitude of a 3D vector

Find the magnitude of $\mathbf{v} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$ .

Apply the 3D magnitude formula
$|\mathbf{v}| = \sqrt{2^2 + (-3)^2 + 6^2}$ .
Compute the squares
$= \sqrt{4 + 9 + 36} = \sqrt{49}$ .
$|\mathbf{v}| = \sqrt{4 + 9 + 36} = \sqrt{49}$
Simplify
$= 7$ .

Result: $|\mathbf{v}| = 7$ .

Typical mistakes

Adding components without the square root when finding a magnitude.
Treating a vector's position in the plane as significant (only magnitude and direction matter).

Active revision

Given $\mathbf{a} = 5\mathbf{i} - 12\mathbf{j}$ , find $|\mathbf{a}|$ and the angle $\mathbf{a}$ makes with the positive $x$ -axis.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Vector arithmetic and unit vectors#

StandardDfE GCE Mathematics J2AQA 7357 J

Triangle law of addition

Fig. 2Placing

\mathbf{b}

head-to-tail after

\mathbf{a}

, the resultant

\mathbf{a} + \mathbf{b}

runs from the start to the finish.

Vectors are added and subtracted component by component:

\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}

. Geometrically, addition follows the triangle law (place the tail of the second at the head of the first; the sum runs from the start to the finish) or, equivalently, the parallelogram law. Subtraction

\mathbf{a} - \mathbf{b}

is addition of the negative,

\mathbf{a} + (-\mathbf{b})

Scalar multiplication scales a vector by a number:

k\mathbf{a}

multiplies every component by

k

, giving a vector

|k|

times as long, in the same direction if

k > 0

and the opposite direction if

k < 0

. A key consequence is the test for parallel vectors: two vectors are parallel exactly when one is a scalar multiple of the other,

\mathbf{a} = k\mathbf{b}

A unit vector has magnitude 1 and is used to specify a direction. To find the unit vector in the direction of

\mathbf{a}

, divide by its magnitude:

\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}

. This is used constantly in mechanics to resolve a force or velocity along a given direction, and to write a vector of a required length in a given direction.

These operations combine: a vector of magnitude

10

in the direction of

\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}

10\,\hat{\mathbf{a}} = 10 \times \dfrac{3\mathbf{i} + 4\mathbf{j}}{5} = 6\mathbf{i} + 8\mathbf{j}

. Being fluent in scaling to a unit vector and back out to a required magnitude is a recurring, examinable skill.

\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}

Unit vector

Divide a vector by its magnitude to obtain a vector of length 1 in the same direction.

\mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} = k\mathbf{b}\ \text{for some scalar } k

Test for parallel vectors

Vectors are parallel exactly when one is a scalar multiple of the other.

Worked example

A vector of given magnitude and direction

Find the vector of magnitude 20 in the direction of $\mathbf{d} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ .

Find the magnitude
$|\mathbf{d}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$ .
Unit vector
$\hat{\mathbf{d}} = \dfrac{1}{3}(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ .
Scale to magnitude 20
$20\hat{\mathbf{d}} = \dfrac{20}{3}(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})$ .
$20\hat{\mathbf{d}} = \tfrac{20}{3}(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})$

Result: $\dfrac{40}{3}\mathbf{i} - \dfrac{20}{3}\mathbf{j} + \dfrac{40}{3}\mathbf{k}$ .

Typical mistakes

Forgetting to divide by the magnitude when constructing a unit vector.
Adding a scalar to a vector, or vice versa (only like objects combine).

Active revision

Find the unit vector in the direction of $\mathbf{a} = 8\mathbf{i} - 6\mathbf{j}$ , and hence a vector of magnitude 15 in the same direction.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Position vectors and geometry#

StandardDfE GCE Mathematics J3AQA 7357 J

Position vectors and the midpoint

Fig. 3

M

, the midpoint of

AB

, has position vector

\tfrac{1}{2}(\mathbf{a} + \mathbf{b})

The position vector of a point

A

is its displacement from the origin, written

\mathbf{a} = \overrightarrow{OA}

. Position vectors turn points into vectors, so geometric statements become vector equations. The displacement from

A

B

is then

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

('finish minus start'), which is the single most used relation in vector geometry, and its magnitude

|\mathbf{b} - \mathbf{a}|

is the distance

AB

The midpoint

M

AB

has position vector

\dfrac{1}{2}(\mathbf{a} + \mathbf{b})

, the average of the two position vectors. More generally, the point dividing

AB

in the ratio

\lambda : \mu

has position vector

\dfrac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda + \mu}

. These formulae are used to find specific points on a segment and to prove results about medians and diagonals.

Collinearity — three points lying on a straight line — is proved by showing that two of the displacement vectors between them are parallel and share a common point: if

\overrightarrow{AB} = k\,\overrightarrow{BC}

then

A

B

C

are collinear. This scalar-multiple test, combined with the 'finish minus start' rule, is the standard route through vector proof questions.

Vector methods prove geometric facts elegantly because they avoid coordinates and lengths. To show a quadrilateral is a parallelogram, for instance, you show one pair of opposite sides are equal as vectors (

\overrightarrow{AB} = \overrightarrow{DC}

); to find where diagonals meet, you equate two expressions for the same point. Setting up such vector equations and solving for the unknown scalars is the core technique.

\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, \qquad \overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})

Displacement and midpoint

Displacement is 'finish minus start'; the midpoint is the average of the position vectors.

Worked example

Proving collinearity

Points $A$ , $B$ , $C$ have position vectors $\mathbf{a} = 2\mathbf{i} - \mathbf{j}$ , $\mathbf{b} = 4\mathbf{i} + 3\mathbf{j}$ , $\mathbf{c} = 7\mathbf{i} + 9\mathbf{j}$ . Show they are collinear.

Find two displacements
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} + 4\mathbf{j}$ ; $\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = 3\mathbf{i} + 6\mathbf{j}$ .
Test for a scalar multiple
$\overrightarrow{BC} = \tfrac{3}{2}(2\mathbf{i} + 4\mathbf{j}) = \tfrac{3}{2}\overrightarrow{AB}$ , so the two are parallel.
$\overrightarrow{BC} = \tfrac{3}{2}\,\overrightarrow{AB}$
Conclude
They are parallel and share the point $B$ , so $A$ , $B$ , $C$ are collinear.

Result: Since $\overrightarrow{BC} = \tfrac{3}{2}\overrightarrow{AB}$ with common point $B$ , the points are collinear.

Typical mistakes

Writing $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$ instead of $\mathbf{b} - \mathbf{a}$ .
Concluding two points coincide from parallel vectors alone, without a common point (parallel is not enough for collinearity).

Active revision

The points $A$ , $B$ , $C$ have position vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ , $\mathbf{b} = 4\mathbf{i} + 8\mathbf{j}$ and $\mathbf{c} = 6\mathbf{i} + 12\mathbf{j}$ . Show that $A$ , $B$ , $C$ are collinear.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Vectors in three dimensions and applications#

AdvancedDfE GCE Mathematics J4AQA 7357 J

A position vector in three dimensions

Fig. 4The position vector of

P(3, 2, 4)

has magnitude

\sqrt{3^2 + 2^2 + 4^2} = \sqrt{29}

Three-dimensional vectors extend every earlier idea by adding a

z

-component and the unit vector

\mathbf{k}

, so a point

P

has position vector

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}

. Arithmetic, magnitude (

\sqrt{x^2 + y^2 + z^2}

), unit vectors, midpoints and the collinearity test all work exactly as in two dimensions, simply carrying the third component along.

The distance between two points in space is the magnitude of the displacement between them:

|\overrightarrow{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

. This is the 3D Pythagoras and is used to compute lengths of edges and diagonals of solids, and separations of points, in three-dimensional geometry problems.

Vectors are the natural language of position and displacement in mechanics. If a particle starts at position vector

\mathbf{r}_0

and moves with constant velocity

\mathbf{v}

, its position after time

t

\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}

. Writing motion this way keeps the components independent and connects directly to the kinematics topic.

The applications combine the tools: finding the distance travelled, showing two paths meet, or determining a point that divides a displacement in a given ratio all reduce to setting up a vector equation and solving for scalars. Working confidently in three components, and interpreting the answer geometrically, is the mark of the full A-Level standard here.

|\overrightarrow{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Distance between two points in 3D

The magnitude of the displacement vector, by three-dimensional Pythagoras.

Worked example

Distance and midpoint in 3D

Find the distance between $A(2, -1, 3)$ and $B(5, 3, 3)$ , and the midpoint of $AB$ .

Displacement
$\overrightarrow{AB} = (5 - 2)\mathbf{i} + (3 - (-1))\mathbf{j} + (3 - 3)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j}$ .
Distance
$|\overrightarrow{AB}| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{25} = 5$ .
$|\overrightarrow{AB}| = \sqrt{9 + 16 + 0} = 5$
Midpoint
$\tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \left(\tfrac{7}{2}, 1, 3\right)$ .

Result: Distance $AB = 5$ ; midpoint $\left(\tfrac{7}{2}, 1, 3\right)$ .

Typical mistakes

Dropping the $z$ -component partway through a calculation.
Mixing up coordinates when forming a 3D displacement vector.

Active revision

Points $A(1, 0, 2)$ and $B(4, 6, -4)$ are given. Find $\overrightarrow{AB}$ , the distance $AB$ , and the position vector of the midpoint of $AB$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Vectors

4 sections~9 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

Vectors

Key points

Magnitude of a 3D vector

Key points

A vector of given magnitude and direction

Key points

Proving collinearity

Key points

Distance and midpoint in 3D

Sources

Vectors

Key points

Magnitude of a 3D vector

Key points

A vector of given magnitude and direction

Key points

Proving collinearity

Key points

Distance and midpoint in 3D

Sources