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StudyForALearn · Practise · Understand
IGCSE · Cambridge (CIE) · 0580

Vector addition, subtraction and scalar multiples

Use directed displacements and components to combine or reverse vectors.

Transformations and vectors pathway · E7.2

Extended only.

Before you begin

  • Read column-vector components.
  • Use signed arithmetic.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What is BA in terms of AB?

What you will learn

  • Add and subtract vectors component by component.
  • Interpret scalar multiplication and reversed direction.

Addition follows a head-to-tail journey

For a = (3,−1) and b = (−2,4), a + b = (1,3). Move along a then b and the resultant joins the start to the final position. Swapping the journey order has the same total displacement, even though the intermediate point differs.

A negative vector reverses its direction: −b = (2,−4). Subtraction a − b means a + (−b), giving (5,−5). A directed vector AB points from A to B; BA = −AB. Direction matters, unlike an ordinary undirected length.

Scalars multiply both components

For scalar k, ka multiplies both horizontal and vertical components. Thus 2a = (6,−2), while −2a = (−6,2). Positive scalars preserve direction, negative scalars reverse it, and the magnitude is scaled by |k|. A zero scalar produces the zero vector.

Use arrows or bold lower-case letters on paper to distinguish vectors from numbers. In this text, a and b refer to directed vectors. Expressions such as 2a − b can be simplified either through components or as an algebraic vector expression; only add quantities with the same vector meaning.

Pause and explain

a = (3,−1), b = (−2,4). Find a + b.

Put the idea to work

Worked example

a = (3,−1), b = (−2,4). Find 2a − b.

Show the worked solution
  1. 2a = (6,−2).
  2. Subtract b component by component: (6−(−2),−2−4).
  3. The result is (8,−6).

Answer (8,−6).

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find (2,5) + (−4,1).

Give me a hint

Follow each component's axis.

Compare my reasoning
  1. Horizontal sum is 2 − 4 = −2.
  2. Vertical sum is 5 + 1 = 6.
  3. The resultant is (−2,6).

(−2,6).

Look for these in your work

  • Added components.
  • Preserved negative signs.
2 · Independent

For a = (2,−3), b = (4,1), find 3a − 2b.

Give me a hint

Multiply before subtracting.

Compare my reasoning
  1. 3a = (6,−9), 2b = (8,2).
  2. Subtract to get (−2,−11).
  3. Check by adding 2b back to recover 3a.

(−2,−11).

Look for these in your work

  • Scaled both components.
  • Checked the subtraction.
3 · Transfer

A journey has displacements (5,2), (−1,4) and a final return to its start. Find the return vector.

Give me a hint

The entire closed journey sums to zero.

Compare my reasoning
  1. The first two total (4,6).
  2. The return must be the negative resultant.
  3. Use (−4,−6); all three sum to (0,0).

(−4,−6).

Look for these in your work

  • Recognised the closed-path condition.
  • Reversed the resultant.

Common mistakes

  • Adding a horizontal to a vertical component.
  • Subtracting without reversing the second vector.
  • Treating directed AB as BA.
Recall without your notes

What can you explain now?

What does a negative scalar do to a nonzero vector?

Compare with the explanation

Reverses direction and scales magnitude by its absolute value.

The sign controls direction while the absolute factor controls physical length.

After trying it yourself, choose your next review. This is your self-assessment.

Your review choice appears on Today. Sign in to sync it across devices.

Make it stick

Tomorrow, draw a head-to-tail journey and calculate its return vector.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.

Number contains 19 sequenced lessons; Algebra and graphs 21; Coordinate geometry 10; Geometry 15; Mensuration 13; Trigonometry 13; Transformations and vectors 11; Probability 7; Statistics 13. All 72 syllabus section entries now link to teaching and staged practice. Seven mixed checkpoints sample skills and suggest review lessons. Nine earlier overviews remain available. Full cumulative assessment and human teacher review remain pending. Written lesson practice, charts and constructions are self-checked, not automatically graded. Checkpoints do not save an exam grade or certify mastery.

Check the official syllabus 2025–2027 Open related practice and resources