Mini-Lesson

Vectors

Vectors have magnitude and direction. This mini-lesson covers i and j components, magnitude and direction, unit vectors, position vectors and displacement, resultants, parallel vectors and collinearity, and how vectors are used in mechanics.

Where this sits: Unit AS 1: Pure Mathematics — vectors in two dimensions. ⚠️ Important CCEA difference: CCEA GCE Mathematics uses 2D vectors only (i and j). Unlike AQA, Edexcel and OCR, 3D vectors (i, j, k) are NOT on the CCEA A-level Mathematics specification — so every vector in this lesson has exactly two components.

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.

2D only

Notation and components

CCEA-specific: vectors at CCEA GCE Mathematics are two-dimensional. You will never be asked for a k-component, a 3D magnitude or a 3D scalar product on this specification. Everything below is in the plane.

A vector can be written in component (i, j) form or as a column vector:

a = 3i + 4j = (3, 4)ᵀi is one unit in the x-direction, j is one unit in the y-direction

Add/subtract component by component: (3i + 4j) + (2i − j) = 5i + 3j.
Scalar multiple: 3(2i − j) = 6i − 3j. It has the same direction, three times as long (a negative scalar reverses the direction).
A scalar (like speed or mass) has size only; a vector (velocity, force, displacement) has size and direction.

Magnitude

Magnitude and direction

|ai + bj| = √(a² + b²)direction: the angle θ with the positive x-direction satisfies tan θ = b/a

Worked example — a = 3i − 4j

|a| = √(3² + (−4)²) = √(9 + 16) = √25 = 5

For 3i + 4j the direction is θ = arctan(4/3) = 53.1° above the positive x-direction.

Sketch it. arctan only ever gives an angle in (−90°, 90°), so for vectors pointing left or down you must add or subtract 180° after checking the quadrant.

Calculate

Magnitude

1Find the magnitude of the vector 3i − 4j.

units

Hint: |3i − 4j| = √(3² + (−4)²) = √(9 + 16) = √25.

Calculate

Direction

2Find the angle, in degrees to 1 decimal place, between the vector 3i + 4j and the positive x-direction.

°

Hint: tan θ = 4/3, so θ = arctan(4/3) = 53.13°. Both components are positive, so the vector is in the first quadrant and no adjustment is needed.

Unit vectors

A unit vector has magnitude 1. To find the unit vector in the direction of a, divide a by its own magnitude:

â = a / |a|this is called normalising the vector

Worked example — a = 5i + 12j

|a| = √(25 + 144) = √169 = 13

â = (5i + 12j)/13 = (5/13)i + (12/13)j ≈ 0.385i + 0.923j

Check: √(0.385² + 0.923²) = √(0.148 + 0.852) = 1 ✓

Calculate

Unit vector

3Find the i-component of the unit vector in the direction of 5i + 12j, to 3 decimal places.

Hint: |5i + 12j| = √(25 + 144) = 13, so the unit vector is (5/13)i + (12/13)j and 5 ÷ 13 = 0.3846.

Position vectors

Position vectors and displacement

The position vector of A is OA, measured from the origin. The displacement from A to B is:

AB = OB − OA"destination minus start" — and |AB| is the distance between the points

Worked example — A(2, 3), B(7, 1)

OA = 2i + 3j and OB = 7i + j

AB = (7i + j) − (2i + 3j) = 5i − 2j

|AB| = √(25 + 4) = √29 = 5.39 units

Order matters: BA = −AB. Getting the subtraction backwards flips the direction (though the magnitude is unchanged).

Calculate

Resultant magnitude

4a = 4i + j and b = −2i + 5j. Find the magnitude of a + b, to 3 significant figures.

units

Hint: a + b = (4 − 2)i + (1 + 5)j = 2i + 6j. Magnitude = √(4 + 36) = √40 = 6.3246.

Parallel

Parallel vectors and collinearity

Two vectors are parallel if one is a scalar multiple of the other: b = λa.

Worked example — find k

p = 3i + kj is parallel to q = 12i − 8j.

q = 4p would need 12 = 4(3) ✓ and −8 = 4k ⇒ k = −2

Check: 3i − 2j scaled by 4 gives 12i − 8j ✓

Collinear points: A, B and C are collinear if AB is parallel to BC (they share the point B, so the vectors must line up).

Calculate

Parallel vectors

5The vectors 3i + kj and 12i − 8j are parallel. Find k.

k

Hint: For parallel vectors the components are in the same ratio: 3/12 = k/(−8), so k = −8 × (3/12) = −2.

Quick check

2D at CCEA

?How many components does every vector on the CCEA GCE Mathematics specification have?

Sort it

Sort by magnitude

Tap a vector, then tap the bin holding its magnitude.

5️⃣ Magnitude 5

🔟 Magnitude 10

1️⃣3️⃣ Magnitude 13

Quick check

Scalar or vector?

?Which of these is a vector quantity?

Match it

Match the vector result

Tap the expression on the left, then its value on the right.

Statement

Answer

Quick check

Vectors in mechanics

?A particle has velocity v = 6i + 8j m/s. What is its speed?

Quick check

Collinear points

?A(1, 2), B(3, 6) and C(6, 12). Are they collinear?

Examiner traps

Vector pitfalls at CCEA

⚠️ Never write a k-component. CCEA vectors are 2D only.
Adding magnitudes. |a + b| ≠ |a| + |b|. Add the components first, then take the magnitude.
Direction angles. arctan alone cannot tell which quadrant you are in — sketch the vector and adjust by 180° if needed.
AB = OB − OA — destination minus start. Reversing it flips the direction.
Parallel ≠ equal. Parallel means one is a scalar multiple of the other.

Recap

The big ideas to know

⚠️ CCEA is 2D only: i and j — no k-component appears on this specification

Magnitude: |ai + bj| = √(a² + b²) · Direction: tan θ = b/a (sketch to fix the quadrant)

Unit vector: â = a/|a| — always has magnitude 1

Displacement: AB = OB − OA (destination minus start)

Parallel: b = λa; collinear points follow from parallel vectors sharing a point

Mechanics link: speed = |v|, and forces add as vectors to give the resultant

Vectors are the language of mechanics — resultants of forces and velocities all use the same component method. Press Finish to see your score.

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