Sec 4 E-Math: Vectors in Two Dimensions

Overview

What you’ll learn

Use column-vector notation and directed-segment notation (AB = b − a) correctly.
Find the image of a point or shape under a translation by a given vector.
Add and subtract vectors and multiply a vector by a scalar, in column form.
Find the magnitude of a vector as √(x² + y²).
Use position vectors and express a given vector in terms of two coplanar vectors.
Solve geometric problems with vectors, including midpoints, parallelograms and collinearity.

In 30 seconds

1A column vector (x, y) records a move of x right and y up; it carries size and direction.
2Directed segment AB = b − a: finish minus start, in position vectors.
3A translation by (a, b) maps (x, y) to (x + a, y + b) ; the shape slides without turning or resizing.
4Add and subtract component-wise: (x₁, y₁) ± (x₂, y₂) = (x₁ ± x₂, y₁ ± y₂).
5Scalar multiple k(x, y) = (kx, ky) ; parallel vectors are scalar multiples of each other.
6Magnitude = √(x² + y²) — Pythagoras on the components.
7Position vector OP is just the coordinates of P; OP = p.
8Midpoint of AB (position vectors a, b) has position vector ½(a + b).

Notes · 1 of 5

Vector notation

Column vectors, directed segments, and the finish-minus-start rule.

A vector carries both size and direction — a displacement. In print, bold letters (a, AB) name vectors; by hand, underline them. The column vector (x, y) records a shift of x horizontally and y vertically: (3, −4) means 3 right and 4 down. For a segment from A to B, the vector AB equals b − a — the position vector of the finish minus that of the start.

1A(1, 2) to B(4, 6)
AB = b − a = (4, 6) − (1, 2) = (3, 4): 3 right, 4 up.
2The reverse journey
BA = a − b = (−3, −4) — same length, opposite direction.
3Order matters
AB and BA are never equal; the arrow in the notation points the way.

Notes · 2 of 5

Translation by a vector

Sliding points and shapes without turning them.

A translation moves every point by the same vector: (x, y) maps to (x + a, y + b) under translation by (a, b). The image keeps the shape and size exactly — only position changes. On a diagram, every vertex slides along a matching arrow.

Point (5, 7) under translation (2, −3) → (7, 4).
Triangle (1, 1), (4, 1), (1, 3) under (3, −2) → (4, −1), (7, −1), (4, 1).
Describing a given translation: subtract the object coordinates from the image coordinates.

Notes · 3 of 5

Adding, subtracting and scaling vectors

Component-wise arithmetic, with magnitude by Pythagoras.

Vectors add and subtract component by component: (x₁, y₁) ± (x₂, y₂) = (x₁ ± x₂, y₁ ± y₂). A scalar multiple k(x, y) = (kx, ky) stretches (or flips, for negative k) the vector while keeping its line of action — so two vectors are parallel exactly when one is a scalar multiple of the other.

u = (3, 1) and v = (1, 2)
Operation	Component working	Result
u + v	(3 + 1, 1 + 2)	(4, 3)
u − v	(3 − 1, 1 − 2)	(2, −1)
2u	(6, 2)	(6, 2)
3u − 2v	(9 − 2, 3 − 4)	(7, −1)

The magnitude (length) of (x, y) is √(x² + y²) — Pythagoras on the components. So |(3, 4)| = 5, |(6, 8)| = 10 and |(2, −1)| = √5 ≈ 2.24 (3 s.f.). Magnitude is always positive (or zero for the zero vector (0, 0)).

Notes · 4 of 5

Position vectors

A point’s address relative to the origin — and midpoints for free.

The position vector of a point P is the vector OP from the origin to P — numerically just the coordinates of P, written OP = p. Adding vectors along a route telescopes: OP + PQ = OQ, so AB = b − a falls out immediately.

P(3, −2) has position vector p = (3, −2).
Route: OA + AB = OB, so AB = b − a.
Midpoint: with position vectors a and b, OM = ½(a + b) — for a = (2, 3) and b = (6, 7), OM = (4, 5).

Notes · 5 of 5

Geometric problems with vectors

Two basis vectors, one whole parallelogram: expressing routes and proving facts.

Exam questions pick two vectors (say a and b) and ask you to express every other route in terms of them. Walk each path along grid edges, adding pieces: in parallelogram OABC with OA = a and OC = c, the diagonal OB = a + c, and the midpoint M of AC satisfies OM = ½(a + c). Opposite sides of a parallelogram give equal vectors: CB = a and AB = c.

1Mark the basis vectors on the diagram
a along one side, c along the other.
2Walk each requested route
Travelling against an arrow negates it: BA = −a.
3Simplify
Collect the a’s and c’s: e.g. AC = −a + c = c − a.
4Prove with equality of vectors
AB = DC (both (2, 1)) proves ABCD is a parallelogram — equal vectors are parallel and equal in length.

To solve for unknown scalars in ma + nb = c, write the components as two simultaneous equations (Chapter 5 again) and solve. The component view and the diagram view always agree — use whichever the question favours, and check the other.

Revise

Key terms

The words examiners expect you to use precisely.

Vector: A quantity with magnitude and direction, e.g. a displacement.
Column vector: The notation (x, y) for a move of x horizontally and y vertically.
Directed segment: A vector along a named segment, e.g. AB = b − a (finish minus start).
Translation: A slide of every point by the same vector; shape and size unchanged.
Position vector: The vector from the origin to a point; numerically the point’s coordinates.
Magnitude: The length of a vector: √(x² + y²).

Show 5 more key terms

Scalar multiple: k(x, y) = (kx, ky); multiples of one vector are parallel to it.
Zero vector: (0, 0) — no movement, magnitude 0.
Equal vectors: Same components: parallel, same length, same direction.
Collinear points: Points on one straight line; shown by parallel vectors sharing a point.
Coplanar vectors: Vectors expressible in terms of two basis vectors a and b as ma + nb.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Computing AB as a − b (start minus finish).
The vector points where the segment travels.
Instead: AB = b − a; BA = a − b.
Mistake: Adding magnitudes instead of vectors: |u| + |v| treated as |u + v|.
The triangle law makes |u + v| depend on the angle between them.
Instead: Add components first, then take the magnitude: |(3, 1) + (1, 2)| = |(4, 3)| = 5, not 2.2 + 2.2.

Show 10 more mistakes

Mistake: Multiplying only one component by the scalar: 2(3, 1) = (6, 1).
Every component scales.
Instead: 2(3, 1) = (6, 2).
Mistake: Dropping the negative when subtracting: u − v computed as u + v.
Subtraction adds the negated vector.
Instead: u − v = u + (−v); negate both components of v first.
Mistake: Forgetting the square root in magnitude.
x² + y² is the magnitude squared.
Instead: For (3, 4): magnitude = √(9 + 16) = 5, not 25.
Mistake: Quoting a magnitude as negative.
Magnitude is a length.
Instead: √(x² + y²) ≥ 0 always.
Mistake: Saying parallel vectors must be equal.
Parallel only needs a scalar multiple.
Instead: (2, 1) and (6, 3) are parallel ((6, 3) = 3(2, 1)) but not equal.
Mistake: Translating a shape by turning or resizing it.
Translation only slides.
Instead: Every vertex moves by the same vector; lengths and angles survive.
Mistake: Using AB = a + b.
Confuses the route rule with the diagonal rule.
Instead: AB = b − a; the + rule is for chained routes: OA + AB = OB.
Mistake: Writing the position vector of P(3, −2) as the point (3, −2) without vector context.
Notation must show it is a vector (column form, bold, or underlined).
Instead: OP = (3, −2) written as a column vector in working.
Mistake: Assuming two vectors with equal magnitudes are equal.
Direction matters.
Instead: (3, 4) and (4, 3) both have magnitude 5 but are different vectors.
Mistake: In ma + nb problems, adding a’s to b’s components.
The letters are separate unknowns.
Instead: Split into two equations: one for x-components, one for y-components.

Check yourself

Quick quiz

32 questions with instant explanations. About 5 minutes.

Question 1 of 3

Question 1: current question
Question 2: not answered
Question 3: not answered

Find magnitudes

The magnitude of the vector (3, 4) is

Practise

Topical questions

Exam-style practice in three tiers, from recall to data questions.

7Foundation
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6Challenging

20 questions worth 82 marks, each with a model answer, mark scheme and examiner tips.

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Full papers for this subject. Each school sets its own scope — check your assessment outline.

Exam practice

Practice paper

A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.

Question paper

Chapter 2 Practice Paper

30 marks · 40 min · 4 pages

7 multiple-choice questions and structured questions with answer lines. Chapter 2 (G7 Vectors in Two Dimensions: notation, translation, arithmetic, magnitude, position vectors, geometric problems).

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Mark scheme

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Chapter 2 Practice Paper — mark scheme

Answers · Marking points · 3 pages

Model answers with a mark for each creditworthy point — for marking at home or in class.

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Where this chapter is tested

Each school sets its own scope — check your assessment outline.

Questions

Frequently asked questions

What does Sec 4 E-Math Chapter 2 cover?

It covers syllabus sub-topic G7 of O-Level 4052 / SEC G3 K310 in full: column-vector and directed-segment notation, translations, position vectors, magnitudes, sums and differences, scalar multiples, expressing vectors in terms of two coplanar vectors, and geometric problems including midpoints, parallelograms and collinearity.

What is the single most useful vector fact here?

AB = b − a — finish minus start in position vectors. Every route question is this rule plus addition: A to B to C telescopes to AC = c − a, and the midpoint rule OM = ½(a + b) follows from averaging the two ends.

How do vectors prove a quadrilateral is a parallelogram?

Show one pair of opposite sides gives the same vector, e.g. AB = DC. Equal vectors are parallel and equal in length, which is exactly the parallelogram condition. The same style proves collinearity: parallel vectors that share a point force a straight line.

Why does the magnitude formula look like Pythagoras?

Because the components (x, y) are the legs of a right triangle whose hypotenuse is the vector itself: |(x, y)| = √(x² + y²). Recognising triples like (3, 4) → 5 and (6, 8) → 10 saves time.

When is this chapter usually tested in school?

Vectors appear in Sec 4 Weighted Assessment 1 and Prelim papers as a 5–8 mark structured question: convert between points and vectors, do arithmetic with a and b, then prove a geometric fact. Clean column-vector notation earns presentation marks.

Sec 4 E-Math Vectors in Two Dimensions notes