Cambridge IGCSE Mathematics 0580: Vectors in two dimensions and magnitude of a vector – Study Guide

This study guide teaches subtopics 7.2 (Vectors in two dimensions) and 7.3 (Magnitude of a vector) from topic 7, Transformations and vectors, of the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027. Both subtopics are Extended only. The Core column of the syllabus says “Extended content only” at C7.2 and C7.3, so this work is examined on Paper 2 (non-calculator) and Paper 4 (calculator), not on Core Papers 1 and 3. Core candidates still need translation by a column vector, which sits in C7.1 and is covered in the topic study guide.

Use this page with the Cambridge IGCSE Mathematics hub and the printable 0580 checklist. For the whole of topic 7, including transformations and vector geometry, see the Transformations and vectors study guide.

What these subtopics cover

0580 reference	What you must be able to do	Tier
E7.2.1	Describe a translation using a vector written as a column vector, as AB (with an arrow) or as a	Extended only
E7.2.2	Add and subtract vectors	Extended only
E7.2.3	Multiply a vector by a scalar	Extended only
E7.3	Calculate the magnitude of a column vector with top entry x and bottom entry y as √(x² + y²)	Extended only

Two notes from the syllabus matter for how questions look:

• Notation. Vectors are printed as AB with an arrow above the letters, or as a bold lower-case letter such as a. You cannot write bold by hand, so the usual convention is to underline the letter: a with a line under it.
• Modulus signs. The magnitude of a is written |a|, and the magnitude of AB is written |AB|.

The magnitude formula is not on the List of formulas printed on page 2 of Papers 2 and 4. You must recall it.

In running text this page writes a column vector as “the column vector (x, y)”, x on top; in worked examples it is stacked, as in the exam.

7.2 Describing a translation with a vector

A translation slides every point of a shape the same distance in the same direction. A column vector records that movement:

( x )   x = movement across: positive to the right, negative to the left
( y )   y = movement up/down: positive up, negative down

So the column vector (−2, 5) means “2 left, 5 up”. It does not say where the movement starts. Any two arrows with the same length and the same direction represent the same vector.

Worked example 1: the vector between two points

A translation maps point P(−3, 5) onto point Q(4, −1). Write the translation as a column vector, then write the vector QP.

Work out how far you move in each direction, always finish minus start:

across: 4 − (−3) = 7      (7 to the right)
up:    −1 − 5    = −6     (6 down)

      (  7 )
PQ =  ( −6 )

The vector QP goes the other way, so both signs change:

      ( −7 )
QP =  (  6 )

QP is the negative of PQ: QP = −PQ. Same length, opposite direction.

Worked example 2: applying a translation

Triangle T has vertices A(1, 2), B(4, 2) and C(1, 6). It is translated by the column vector (−5, 3). Find the vertices of the image, and the vector that maps the image back onto T.

Add the vector to each point: subtract 5 from every x-coordinate and add 3 to every y-coordinate.

A(1, 2) → (1 − 5, 2 + 3) = (−4, 5)
B(4, 2) → (4 − 5, 2 + 3) = (−1, 5)
C(1, 6) → (1 − 5, 6 + 3) = (−4, 9)

To undo the translation, reverse it: the column vector (5, −3).

When a question says “describe fully the single transformation”, you need both the word “translation” and the column vector. The vector alone is not a description.

7.2 Adding and subtracting vectors

Adding

Add the top entries, then add the bottom entries. Geometrically, a + b means “do translation a, then do translation b”. The single translation that has the same effect is a + b. In letters, AB + BC = AC: going from A to B and then from B to C gets you from A to C.

Subtracting

Subtract entry by entry. It helps to think of a − b as a + (−b): add the vector that points the opposite way to b. Order matters: a − b and b − a are negatives of each other.

Worked example 3

p is the column vector (5, −2) and q is the column vector (−1, 7). Find p + q, p − q and q − p.

          (  5 )   ( −1 )   (  5 + (−1) )   ( 4 )
p + q  =  ( −2 ) + (  7 ) = ( −2 + 7    ) = ( 5 )

          (  5 )   ( −1 )   (  5 − (−1) )   (  6 )
p − q  =  ( −2 ) − (  7 ) = ( −2 − 7    ) = ( −9 )

          ( −1 )   (  5 )   ( −1 − 5    )   ( −6 )
q − p  =  (  7 ) − ( −2 ) = (  7 − (−2) ) = (  9 )

Check: q − p is exactly −(p − q), as it must be.

Sign errors happen on “minus a negative”. Write the bracket in, as above, before you simplify.

7.2 Multiplying a vector by a scalar

A scalar is an ordinary number. To multiply a vector by a scalar k, multiply both entries by k.

What k does to the arrow:

Value of k	Effect on the vector
k > 1	same direction, longer
0 < k < 1	same direction, shorter
k = −1	same length, opposite direction
k < 0	opposite direction, length multiplied by the size of k

In every case ka is parallel to a. Showing that one vector is a scalar multiple of another is the key move in vector geometry proofs, which belong to E7.4.

Worked example 4

r is the column vector (6, −4). Find 3r, −2r and ½r. Then, with p and q from Worked example 3, find 3p − 2q.

3r   = (3 × 6,  3 × (−4))  = (18, −12)
−2r  = (−2 × 6, −2 × (−4)) = (−12, 8)
½r   = (½ × 6,  ½ × (−4))  = (3, −2)

(Each answer above is a column vector, written across to save space.)

For a combination, scale first, then add or subtract:

         (  15 )   ( −2 )   ( 15 − (−2) )   (  17 )
3p − 2q = ( −6 ) − ( 14 ) = ( −6 − 14   ) = ( −20 )

Worked example 5: finding unknowns

Find x and y if the column vector (x, 4) plus 3 times the column vector (2, y) equals the column vector (11, −5).

Two vectors are equal only when their top entries are equal and their bottom entries are equal. So split the vector equation into two ordinary equations:

top:     x + 3 × 2 = 11   →  x + 6 = 11   →  x = 5
bottom:  4 + 3y    = −5   →  3y = −9      →  y = −3

x = 5, y = −3.

Worked example 6: writing one vector in terms of two others

a is the column vector (2, 1) and b is the column vector (−1, 3). Find the numbers m and n such that ma + nb is the column vector (1, 11).

Compare top entries and bottom entries:

top:     2m − n  = 1     ... (1)
bottom:  m + 3n  = 11    ... (2)

From (1), n = 2m − 1. Substitute into (2):

m + 3(2m − 1) = 11
m + 6m − 3    = 11
7m            = 14   →  m = 2
n = 2(2) − 1  = 3

m = 2, n = 3. Check: 2a + 3b = (4, 2) + (−3, 9) = (1, 11). Correct.

This is a pair of simultaneous equations (E2.5). Always substitute back as a check.

7.3 Magnitude of a vector

The magnitude of a vector is its length. For the column vector (x, y), the arrow is the hypotenuse of a right-angled triangle with horizontal side x and vertical side y. Pythagoras’ theorem gives:

|a| = √(x² + y²)

Squaring removes any minus sign, so a magnitude is never negative. The direction of the vector makes no difference: |AB| = |BA|.

Worked example 7 (non-calculator)

Find the magnitude of the column vector (−8, 15).

|v| = √((−8)² + 15²)
    = √(64 + 225)
    = √289
    = 17

Put a negative entry in brackets before squaring. Without them, −8² can be misread as −64.

Worked example 8: exact answer in surd form (non-calculator)

Find the exact magnitude of the column vector (4, −6).

|v| = √(4² + (−6)²) = √(16 + 36) = √52
√52 = √(4 × 13) = 2√13

2√13. On Paper 2, “exact” means leave the surd; simplifying surds is part of E1.18.

Worked example 9 (calculator)

A translation maps A(1, −4) onto B(6, 3). Find |AB|.

       ( 6 − 1    )   ( 5 )
AB  =  ( 3 − (−4) ) = ( 7 )

|AB| = √(5² + 7²) = √74 = 8.602325...

|AB| = 8.60 (3 s.f.). The syllabus asks for non-exact answers to 3 significant figures unless the question says otherwise, so keep the final zero: 8.6 is only 2 significant figures.

Finding |AB| from coordinates is the same calculation as the length of a line segment in E3.4.

Worked example 10: magnitude of a combination

a is the column vector (3, −1) and b is the column vector (−2, 4). Find |2a − b| and compare |a + b| with |a| + |b|.

Find the vector first, then its magnitude:

2a − b = (6, −2) − (−2, 4) = (8, −6)
|2a − b| = √(8² + (−6)²) = √100 = 10

a + b = (1, 3)
|a + b| = √(1² + 3²) = √10 = 3.16 (3 s.f.)

|a| = √10 = 3.162...,  |b| = √20 = 4.472...
|a| + |b| = 7.63 (3 s.f.)

So |a + b| is not |a| + |b|. Never work out magnitudes separately and add them.

Worked example 11: an unknown entry

The column vector (k, −24) has magnitude 25. Find the possible values of k.

k² + (−24)² = 25²
k² + 576    = 625
k²          = 49
k           = 7 or k = −7

Both values work, because squaring hides the sign. Give both unless the question restricts k.

Worked example 12: a vector with a given magnitude

Find a vector of magnitude 25 that is parallel to the column vector (8, −6).

|(8, −6)| = √(64 + 36) = √100 = 10
25 ÷ 10 = 2.5
2.5 × (8, −6) = (20, −15)

The column vector (20, −15) works, and so does (−20, 15), which points the opposite way. This combines E7.2 (scalar multiples are parallel) with E7.3 (magnitude).

Paper 2 and Paper 4

Both Extended papers can test this content.

• Paper 2 (non-calculator): expect whole-number arithmetic with negatives, Pythagorean triples such as 8, 15, 17, and exact surd answers. Practise the squares up to 25² so the arithmetic is quick.
• Paper 4 (calculator): magnitudes are often not exact. Write the unrounded root (√74) before the decimal, and round only at the end. If you use a magnitude later in the question, use the unrounded value.

Common errors

• Working out a translation vector as start minus finish. It is finish minus start; check one point by adding the vector back.
• Losing a sign on “minus a negative” in subtraction, especially in 3p − 2q type questions.
• Multiplying only the top entry by the scalar.
• Forgetting the square root, so giving 289 instead of 17.
• Squaring a negative entry without brackets and getting −64 instead of 64.
• Adding magnitudes instead of adding vectors first.
• Giving only the positive root when the unknown could be negative.
• Rounding √74 to 8.6 instead of 8.60.

Where this leads

E7.4 Vector geometry (Extended only) builds directly on this page. You represent vectors as directed line segments, use position vectors, write routes through a diagram in terms of two given vectors, and use scalar multiples to show that lines are parallel or that three points are collinear. The Transformations and vectors study guide has a worked parallel-lines proof.

Recap the whole topic with the Transformations and vectors revision notes and test yourself with the Transformations and vectors practice questions. If the Pythagoras step feels shaky, the Pythagoras’ theorem study guide covers it, and the Coordinate geometry study guide covers length and midpoint. Check where you stand with the Extended diagnostic (or the Core diagnostic if you sit Core).

Official syllabus

Cambridge International, Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027 (Version 3), Subject content, topic 7 Transformations and vectors: E7.2 Vectors in two dimensions and E7.3 Magnitude of a vector (Extended only; C7.2 and C7.3 are marked “Extended content only”). Linked references: C7.1/E7.1 Transformations, E7.4 Vector geometry, E1.18 Surds, E2.5 Equations and E3.4 Length and midpoint.