G24: Vectors
Describe translations as 2D vectors; apply addition and subtraction of vectors; scalar multiplication
📋 Key Concepts
A vector describes a movement in terms of direction and magnitude. It is written as a column vector.
Vector =
| x |
| y |
Where x is the horizontal movement (positive = right) and y is the vertical movement (positive = up).
📝 Column Vector Notation
Writing vectors: The top number is horizontal movement, the bottom number is vertical movement.
Example 1
The vector
| 3 |
| 2 |
The vector
| -4 |
| 1 |
The vector
| 2 |
| -5 |
📝 Vector Addition
Adding vectors: Add the corresponding components.
| a |
| b |
| c |
| d |
| a+c |
| b+d |
Example 2
Find
| 3 |
| 2 |
| 1 |
| 4 |
Solution:
| 3 |
| 2 |
| 1 |
| 4 |
| 3+1 |
| 2+4 |
| 4 |
| 6 |
📝 Vector Subtraction
Subtracting vectors: Subtract the corresponding components.
| a |
| b |
| c |
| d |
| a-c |
| b-d |
Example 3
Find
| 5 |
| 3 |
| 2 |
| 7 |
Solution:
| 5 |
| 3 |
| 2 |
| 7 |
| 5-2 |
| 3-7 |
| 3 |
| -4 |
📝 Scalar Multiplication
Multiplying by a scalar (number): Multiply each component by the scalar.
k
=
| a |
| b |
| ka |
| kb |
Example 4
Find 3
| 2 |
| -1 |
Solution:
3
| 2 |
| -1 |
| 6 |
| -3 |
Example 5
Find -2
| 4 |
| -5 |
Solution:
-2
| 4 |
| -5 |
| -8 |
| 10 |
📝 Combined Operations
Example 6
Find 2
| 3 |
| 1 |
| -1 |
| 4 |
Solution:
=
| 6 |
| 2 |
| -1 |
| 4 |
=
| 5 |
| 6 |
Example 7
Find 3
| a |
| 2 |
| 2 |
| b |
| 1 |
| 3 |
| 4 |
| 1 |
Solution:
3
| 1 |
| 3 |
| 4 |
| 1 |
=
| 3 |
| 9 |
| 4 |
| 1 |
=
| -1 |
| 8 |
📝 Vector Geometry
Vector between two points: If A is at (x₁, y₁) and B is at (x₂, y₂), then:
Vector AB =
| x₂-x₁ |
| y₂-y₁ |
Example 8
Find the vector from A(2, 3) to B(7, 1).
Solution:
AB =
| 7-2 |
| 1-3 |
| 5 |
| -2 |
📝 Parallel Vectors
Parallel vectors: Two vectors are parallel if one is a scalar multiple of the other.
Example 9
Are
| 2 |
| 4 |
| 3 |
| 6 |
Solution:
| 3 |
| 6 |
| 2 |
| 4 |
Yes, they are parallel (same direction, different magnitude).
🧮 Worked Examples — Exam Style
Column vectors add/subtract component by component. The journey rule (A to C via B) and the scalar-multiple parallel test are the two exam favourites.
Given AB = (3, −1) as a column vector and BC = (2, 4), find AC.
Journey rule: A → C = A → B then B → C
AC = AB + BC = (3 + 2, −1 + 4) = (5, 3)
M1 for adding the components, A1 for (5, 3). Column vectors are written vertically in the exam — practise the layout.
Is the vector (6, 9) parallel to the vector (2, 3)? Explain.
Parallel vectors are scalar multiples of each other.
(6, 9) = 3 × (2, 3) — one multiplier works for BOTH components → yes, parallel
B1 yes with B1 justification (×3). Both components must scale by the same factor.
Exam wording to expect: "Write (vector) as a column vector", "Find AB as a column vector", "Show that … is parallel to …".
❓ Practice Questions
Q1: Find
| 5 |
| -2 |
| 3 |
| 7 |
Q2: Find 4
| 2 |
| -3 |
Q3: Find
| 8 |
| 5 |
| 3 |
| 9 |
Q4: Find 2
| 4 |
| 1 |
| -2 |
| 5 |
Q5: Find the vector from point P(1, 4) to Q(6, 2).
Q6: Given AB = (3, −1) and BC = (2, 4) as column vectors, find AC
Q7: Is (6, 9) parallel to (2, 3)? Explain.
Q6: Given AB = (3, −1) and BC = (2, 4) as column vectors, find AC
Q7: Is (6, 9) parallel to (2, 3)? Explain.
Q6: Given AB = (3, −1) and BC = (2, 4) as column vectors, find AC
Q7: Is (6, 9) parallel to (2, 3)? Explain.
✅ Answers
8 5 8 -12 5 -4
+8 2
=-6 15 2 17 5 -2
🧮 Worked Examples — Harder Exam Style
OP = (4, −3) and OQ = (−2, 5) as position vectors (from O). Find PQ as a column vector, and state whether PQ is parallel to (6, −8).
PQ = Q − O − (P − O) = OQ − OP = (−2 − 4, 5 − (−3)) = (−6, 8)
Parallel test: is (−6, 8) = k × (6, −8)? k = −1 works for both components → yes, parallel (PQ points the opposite way along the same direction).
M1 for OQ − OP, A1 for (−6, 8), B1 for parallel with k = −1. A negative scalar still means parallel — same line direction, opposite sense.
📋 Model Answers — Full Working
Every mark in GCSE maths comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. Add component by component: 5 + 3 = 8, −2 + 7 = 5 → (8, 5)
Q2. Scalar multiply each component: 4 × 2 = 8, 4 × (−3) = −12 → (8, −12)
Q3. Subtract component by component: 8 − 3 = 5, 5 − 9 = −4 → (5, −4)
Q4. 2 × (4, 1) = (8, 2) and 3 × (−2, 5) = (−6, 15); add: (8 − 6, 2 + 15) = (2, 17)
Q5. PQ = Q − P = (6 − 1, 2 − 4) = (5, −2). End point minus start point — order matters and reversing gives the negative.
Q6. AC = (5, 3) — journey rule AB + BC
Q7. Yes: (6, 9) = 3 × (2, 3), same scalar for both components
📌 Key Facts to Memorise
- Column vector (a above b): a = right (+) / left (−), b = up (+) / down (−)
- Add/subtract component by component; scalar multiply BOTH components
- Journey rule: AC = AB + BC (any route between the same two points is equal)
- Vector from P to Q = Q − P (end minus start)
- Parallel vectors: one is a scalar multiple of the other (both components, same factor)
🗺️ In the Exam
Column-vector arithmetic: 2–3 marks on Paper 1 (non-calculator). The parallel test and journey rule add the AO2 marks. Vectors reappear in transformations (translation by a vector).
Links to: translation (G), coordinates (A8), midpoint and line segments (A).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Remember: top is horizontal (x), bottom is vertical (y)
- For scalar multiplication, multiply both components
- When adding/subtracting, work with each row separately
- Negative scalar reverses the direction
- Parallel vectors have the same ratio between components
- Draw vectors on a diagram to visualise them
🧠 Problem-Solving Strategies
Problem-Solving
Vectors have magnitude and direction. Column vectors: (a, b) means a right and b up. Add vectors by adding components. Subtract by subtracting components. Multiply by a scalar: k(a, b) = (ka, kb). The magnitude of (a, b) = sqrt(a squared + b squared). Parallel vectors are scalar multiples of each other.
Multi-Step Problem
Vector a = (3, 4) and vector b = (1, -2). Find: (i) 2a + b, (ii) the magnitude of 2a + b, (iii) a unit vector in the direction of a.
Solution: (i) 2a = (6, 8), so 2a + b = (7, 6). (ii) Magnitude = sqrt(49 + 36) = sqrt(85) = 9.22. (iii) Magnitude of a = sqrt(9 + 16) = 5. Unit vector = (3/5, 4/5) = (0.6, 0.8).
⚠️ Common Errors
Watch Out!
1. Wrong: Adding components incorrectly: (3,4) + (1,2) = (4,2) Correct: Add corresponding components: (3,4) + (1,2) = (3+1, 4+2) = (4, 6). The x-components add and the y-components add separately.
2. Wrong: Calculating magnitude as a + b: magnitude of (3,4) = 3 + 4 = 7 Correct: Magnitude = sqrt(3 squared + 4 squared) = sqrt(9 + 16) = sqrt(25) = 5. Use Pythagoras, not addition.
3. Wrong: Thinking parallel vectors must have the same direction Correct: Parallel vectors are scalar multiples of each other. If a = 2b, they are parallel and same direction. If a = -2b, they are parallel but opposite direction. Both cases are parallel.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: Given that vector a = (2, 3) and vector b = (4, -1): (a) Find 3a - 2b. (b) Find the magnitude of a + b. (c) Find a scalar k such that ka + b is parallel to (1, 5).
(a) 3a = (6, 9), 2b = (8, -2). 3a - 2b = (6-8, 9-(-2)) = (-2, 11).
(b) a + b = (6, 2). Magnitude = sqrt(36 + 4) = sqrt(40) = 2 x root10 = 6.32.
(c) ka + b = (2k + 4, 3k - 1). For parallel to (1, 5): (2k + 4)/(3k - 1) = 1/5. 5(2k + 4) = 3k - 1. 10k + 20 = 3k - 1. 7k = -21. k = -3.
Mark scheme: M1 scalar multiplication, A1 (-2, 11), M1 magnitude, A1 2 x root10, M1 parallel condition, A1 k = -3
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A hiker walks 3 km East then 4 km North. They then walk back to the start.
(a) Express each leg of the journey as a vector.
(b) What is the total displacement after the first two legs?
(c) A student says "The total distance walked is the same as the magnitude of the total displacement." Is this correct for the first two legs? Explain.
Answers: (a) First leg: (3, 0). Second leg: (0, 4). Return: (-3, -4). (b) Total displacement = (3, 4). Magnitude = 5 km, direction = arctan(4/3) = 53.1 degrees from East. (c) No — total distance = 3 + 4 = 7 km. Total displacement magnitude = 5 km. Distance is the total path length; displacement is the straight-line distance from start to finish. They are only equal if the path is a straight line.
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