Vectors — AQA GCSE Mathematics

SPEC G24, G25 GEOMETRY & MEASURES

F+H ALL PAPERS

Vectors — Notation & Arithmetic

Column vector form, addition, subtraction, scalar multiplication, magnitude, and the relationship between position vectors and the vector between two points.

What you will learn

By the end of this lesson you will:

Read and write column vectors and recognise the bold-letter and arrow notations

Draw a column vector on a coordinate grid and interpret it as a translation

Add and subtract vectors component-wise (and visualise the head-to-tail rule)

Apply scalar multiplication : k times a vector scales each component by k

Find the magnitude of a vector using Pythagoras: |v| = (x² + y²)

Use position vectors and the rule AB = b − a (destination minus start)

Recognise when two vectors are parallel (one is a scalar multiple of the other)

Higher students — preview

Everything you learn here is the foundation for vector geometric proofs in Lesson 4.20. The arithmetic is the same; the difference is what you do with it.

What is a vector?

A vector has two pieces of information: a magnitude (how big / how long) and a direction (which way). This contrasts with a scalar, which only has size (e.g. mass, temperature, time).

Vector notations you'll see

The same vector can be written in three equivalent ways:

Column form:(xy) — the most common form at GCSE

Bold lower-case letter:a, b — in printed text. Hand-written, you underline: a

Arrow notation:AB means "the vector going from point A to point B"

Vectors are not points

A column vector (32) means "3 right, 2 up" — it's a displacement, not a position. The same vector can sit anywhere on the grid; what matters is how far and in which direction it moves you.

Column Vectors on a Grid

The column vector (xy) tells you to move x units horizontally (right if positive, left if negative) and y units vertically (up if positive, down if negative).

a = (32) means "3 right, 2 up"

Sign conventions

Negative x = left. Negative y = down. So (−4−1) means "4 left, 1 down".

Adding Vectors

To add two column vectors, add their components separately:

(ab) + (cd) = (a + cb + d)

a = (21), b = (13), a + b = (34)

Geometric meaning — head-to-tail rule

Place the tail of b at the head of a. The vector from start to finish is a + b — the dashed red arrow above. Same answer either way: arithmetic or visual.

Subtracting Vectors

Subtraction works the same way — subtract components separately:

(ab) − (cd) = (a − cb − d)

Subtraction = adding the reverse

Geometrically, a − b means "a + (−b)". To draw it: reverse the direction of b first, then place head-to-tail with a.

If b = (13) then − b = (−1−3) — same magnitude, reversed direction.

Worked example — quick

If a = (52) and b = (3−1), then a − b = (5 − 32 − (−1)) = (23). Watch the double negative: 2 − (−1) = 2 + 1 = 3.

Scalar Multiplication

A scalar is just a number. Multiplying a vector by a scalar k stretches each component by k:

k (ab) = (k ak b)

Effect of k

k > 1: same direction, longer

0 < k < 1: same direction, shorter

k < 0: reversed direction, magnitude scales by |k|

k = 0: zero vector (no displacement)

Quick check

3 (2−5) = (6−15)

Both components multiplied by 3.

Magnitude (Length) of a Vector

The magnitude of a vector is its length — how far the displacement actually is. For a column vector, it's just Pythagoras on the components:

| (xy) | = (x² + y²)

Worked computation

For v = (34): |v| = (3² + 4²) = (9 + 16) = 25 = 5The vector's length is 5 units — you'd recognise this as the hypotenuse of a 3-4-5 right triangle. Negative components don't change the magnitude (they get squared).

01Worked Example — Arithmetic & Magnitude

Given a = ( 4 −3 ) and b = ( −1 5 ) :

(a) Find 2 a + b as a column vector. [2 marks]

(b) Find | a |. [2 marks]

M — Method

(a) Multiply each component of a by 2, then add to b component-wise.

(b) Apply Pythagoras to the components of a: |a| = (x² + y²).

C — Calculate

(a)

2a = 2 (4−3) = (8−6)

2a + b = (8 + (−1)−6 + 5) = (7−1)

(b)

|a| = (4² + (−3)²) = (16 + 9) = 25 = 5

E — Evaluate

(a) 2 a + b = ( 7 −1 ) (b) | a | = 5

The 4-3-5 right-angled triple shows up naturally when you compute the magnitude. Negative components squared become positive — (−3)² = 9, not −9.

Position Vectors

A position vector is the vector from the origin O to a point. If A has coordinates (3, 2), then the position vector of A is

_OA = a = (32)

Notation convention

The bold lowercase letter matches the capital point label: point A has position vector a; point B has position vector b. This convention is used everywhere in vector geometry.

For any point P with coordinates (x, y), the position vector is just

_OP = (xy)

Why this matters

Position vectors connect points (locations on a grid) to vectors (displacements). Once you can switch between the two, you can find the vector between any two points using a simple subtraction — that's the next slide.

The Vector AB

The vector from point A to point B is

AB = b − a

(destination minus start)

Why "b minus a"?

To go from A to B, you can first walk backwards along a (returning to O), then forwards along b (out to B). That's − a + b = b − a.

Top examiner-flagged error

Students often write AB = a − b. Wrong — that gives you _BA, the vector going the other way. The arrow A→B reads "b minus a" (destination first).

02Worked Example — Vector Between Two Points

Point A has coordinates (2, 1) and point B has coordinates (5, 5).

(a) Write AB as a column vector. [2 marks]

(b) Find the magnitude | AB |. [2 marks]

M — Method

(a) The position vectors are a = (21) and b = (55). Apply AB = b − a.

(b) Apply Pythagoras to the components of AB.

C — Calculate

(a)

AB = b − a = (5 − 25 − 1) = (34)

(b)

|AB| = (3² + 4²) = (9 + 16) = 25 = 5

E — Evaluate

(a) AB = ( 3 4 ) (b) | AB | = 5

|AB| is the distance between A and B — in this case, 5 units. This is also exactly what you'd get from the distance-between-two-points formula.

Parallel Vectors

Two vectors are parallel if one is a scalar multiple of the other:

ab ↔ a = k b for some scalar k

Common misconception — opposite still counts

"Parallel" does NOT mean "same direction". If a = kb with negative k, the vectors point in opposite directions but are still parallel by the definition. Only vectors that are not scalar multiples at all are non-parallel.

How to test for parallel

Given u = (x₁y₁) and v = (x₂y₂): find k from the x-components (k = x₂ / x₁), then check the y-components match: y₂ should equal k y₁. If both match, they're parallel; if not, they aren't.

03Worked Example — Testing for Parallel

Determine, in each case, whether the two vectors are parallel. Justify your answer.

(a) u = ( 3 −2 ) and v = ( −9 6 ) . [2 marks]

(b) p = ( 1 2 ) and q = ( 3 5 ) . [2 marks]

For each pair, find the value of k from the x-components, then check whether the y-components also satisfy that k. Match → parallel; mismatch → not parallel.

C — Calculate

(a) If v = ku, from the x-components: −9 = k(3) → k = −3. Check y-components: −3 × (−2) = 6 ✓ matches.

(b) If q = kp, from the x-components: 3 = k(1) → k = 3. Check y-components: 3 × 2 = 6, but the y-component of q is 5. Mismatch.

E — Evaluate

(a) u and v ARE parallel (k = −3 — opposite direction, but still parallel by definition).

(b) p and q are NOT parallel — no single k makes both components match.

Examiner Alerts

Alert 1 — Wrong order on AB

Students compute AB as a − b instead of b − a. What to write instead: read the arrow direction A → B as "destination minus start", so AB = b − a. The wrong order gives the right magnitude but the opposite direction — you'd lose the accuracy mark.

Alert 2 — Parallel doesn't mean "same direction"

Students reject a pair as "not parallel" because the vectors point opposite ways. What to write instead: parallel just means one is a scalar multiple of the other — k can be negative. Always test by computing k from x-components and checking against y-components.

Alert 3 — Sign errors with column vector subtraction

When a component is itself negative, students forget the double-negative: 5 − (−3) = 8, not 2. What to write instead: always write the substitution explicitly with brackets — e.g. (5) − (−3) — before evaluating.

Common Errors

Students often writeCorrect versionWhy it matters

AB = a − bAB = b − aDestination minus start. The wrong way gives _BA.

| a | = a + b for a = (a, b)| a | = (a² + b²)Magnitude uses Pythagoras, not addition.

Parallel = pointing the same wayParallel = scalar multiple (k can be negative)−2 a is parallel to a .

2 ( 3 4 ) = ( 6 4 )2 ( 3 4 ) = ( 6 8 )Multiply both components by the scalar.

5 − (−3) = 25 − (−3) = 8Subtracting a negative is adding.

| a | = −5 (for negative components)Magnitude is always non-negativeNegative components squared become positive.

Practice

(a) Point A has position vector a = ( 4 1 ) and point B has position vector b = ( −2 9 ) . Find AB as a column vector and find | AB |. [4 marks]

(b) Determine whether c = ( 3 −5 ) and d = ( −9 15 ) are parallel. Justify your answer. [2 marks]

M — Method

(a) Apply AB = b − a; then magnitude via Pythagoras.

(b) Find k from x-components, check y-components.

C — Calculate

(a)

AB = b − a = (−2 − 49 − 1) = (−68)

|AB| = ((−6)² + 8²) = (36 + 64) = 100 = 10

(b) From x-components: −9 = k(3) → k = −3. Check y-components: −3 × (−5) = 15 ✓ matches.

E — Evaluate

(a) AB = ( −6 8 ) , | AB | = 10.

(b) c and d are parallel — d = −3c (a scalar multiple, with negative scalar meaning opposite direction).

The 6-8-10 triple in part (a) is just a scaled 3-4-5.

Lesson checkpoint — you should now be able to:

Read and write column vectors and use bold-letter and arrow notation

Add, subtract, and scalar-multiply column vectors component-wise

Compute the magnitude of a vector using Pythagoras

Find AB from position vectors using b − a

Test whether two vectors are parallel (one is a scalar multiple of the other — sign of k may be either)

Where This Leads ⬣ HIGHER PREVIEW

⬣ Higher only — Lesson 4.20

All the arithmetic from this lesson becomes a tool for vector geometric proofs at Higher tier (spec ref G25 Higher). Typical questions:

Prove that two lines are parallel — show that one direction vector is a scalar multiple of the other

Prove three points are collinear — show that AB = k _BC for some scalar k (which forces all three points onto the same straight line)

Prove a midpoint or ratio property — express the relevant vectors in terms of base vectors a and b and compare

The strategy is always the same: express every vector you need in terms of the given base vectors, then compare. Vector proofs typically appear at the back of Higher papers and discriminate at grade 8–9 — mastering them is high-value.

What every student needs from this lesson

Whether you're sitting Foundation or Higher, the F+H content above is the full requirement: column vectors, arithmetic, magnitude, AB = b − a, parallel test. Master these and you've covered all of G24 and the F+H portion of G25. Higher tier students continue to 4.20 for proofs.