A-Level · Mathematics · AQA · Mark scheme decoded
AQA A-Level Mathematics: Position Vectors and Distance Between Points — mark scheme explained
The short answer
In AQA A-Level Mathematics, understanding position vectors and calculating the distance between two points represented by these vectors is a crucial skill. This section will cover what position vectors are, how to represent them, and the method for finding the distance between two points using vector notation. What Are Position Vectors?
The question
Point A has coordinates (1, 3) and point B has coordinates (4, 7). Calculate the distance between points A and B.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For questions involving position vectors and distance, marks are typically awarded for correctly identifying the position vectors, performing the subtraction to find AB , and calculating the magnitude. Ensure each step is clearly shown.
What the command words demand
- Calculate
- Perform the necessary steps to find the distance between two points using position vectors.
- Determine
- Identify and use the correct formula to find the magnitude of a vector.
- Find
- Locate the required information and apply it to solve the problem.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 3-5 minutes per question to ensure you have enough time to accurately perform calculations and show all steps.
- Find the position vectors OA and OB :0 marks
- OA = 1 i + 3 j0 marks
- OB = 4 i + 7 j0 marks
- Calculate the vector AB :1 mark
- AB = (4 - 1) i + (7 - 3) j = 3 i + 4 j1 mark
- Find the magnitude of AB :0 marks
- | AB | = √(3 2 + 4 2 ) = √(9 + 16) = √25 = 5 units2 marks
Final answer: 5 units
Work through every step correctly and you earn all 4 marks.
Another worked example
Point A has coordinates (0, 0, 0) and point B has coordinates (3, 4, 5). Calculate the distance between points A and B.
- Find the position vectors OA and OB :0 marks
- OA = 0 i + 0 j + 0 k1 mark
- OB = 3 i + 4 j + 5 k0 marks
- Calculate the vector AB :1 mark
- AB = (3 - 0) i + (4 - 0) j + (5 - 0) k = 3 i + 4 j + 5 k1 mark
- Find the magnitude of AB :0 marks
- | AB | = √(3 2 + 4 2 + 5 2 ) = √(9 + 16 + 25) = √50 ≈ 7.07 units2 marks
Final answer: 7.07 units
Work through every step correctly and you earn all 5 marks.
Common mistakes
Forgetting to subtract the position vectors correctly when finding AB
Why it happens: Students sometimes confuse the order of subtraction, leading to incorrect vector components.
Fix: Always ensure that you subtract the coordinates of point A from those of point B: AB = OB - OA .
Using the wrong formula for the magnitude of a vector
Why it happens: Students might use the Pythagorean theorem without considering all components in three-dimensional space.
Fix: Use the correct formula for the magnitude of a vector: | v | = √( a 2 + b 2 + c 2 ) in three dimensions.
Forgetting to take the square root when calculating the magnitude
Why it happens: Students might stop at the sum of squares, forgetting the final step.
Fix: Always remember to take the square root after squaring and summing the components: | v | = √( a 2 + b 2 + c 2 ).
Confusing the position vector with the coordinates of a point
Why it happens: Students might write the coordinates directly without using unit vectors.
Fix: Always represent the position vector using unit vectors: OP = x i + y j + z k .
Using the wrong components when calculating the magnitude
Why it happens: Students might mix up the x, y, and z components when substituting into the formula.
Fix: Double-check that you are using the correct components for each term in the formula: | v | = √( a 2 + b 2 + c 2 ).
Forgetting to include all components in three-dimensional space
Why it happens: Students might overlook the z-component when working in three dimensions.
Fix: Ensure that you include all components (x, y, and z) when calculating the magnitude of a vector in three-dimensional space.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Distance Between Points | Find the distance between two points using their position vectors and magnitude. | 4 |
| Distance Between Points | Find the distance between two 3D points using position vectors and magnitude. | 5 |
| Distance Between Points | Find the distance between two coordinate points using the distance formula. | 5 |
| 3D Distance Between Points | Find the distance between two 3D points using their coordinates and the magnitude of the vector. | 5 |
| Total across these question types | 19 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.