PassNova has 18 free A-level Maths practice questions on Vectors, each with a clear explanation. Practise them in the browser with instant feedback — 100% free, no sign-up, on any device. Updated for 2026.
Vectors: example questions & answers
18 worked examples with answers and explanations below. Practise them in the browser with instant feedback on every answer.
Find the magnitude of the vector 5i − 12j.
- A17
- B7
- C13✓
- D169
Answer: The magnitude is √(5² + (−12)²) = √(25 + 144) = √169 = 13. Adding the components gives 5 + 12 = 17, and omitting the final square root leaves 169.
Find the unit vector in the direction of 6i + 8j.
- A0.6i + 0.8j✓
- B3i + 4j
- C6i + 8j
- D0.6i − 0.8j
Answer: The magnitude is √(6² + 8²) = √100 = 10, so the unit vector is (6i + 8j)/10 = 0.6i + 0.8j. Dividing by 2 instead of by the magnitude 10 gives 3i + 4j, which is not a unit vector.
Given a = 2i + 3j and b = 4i − j, find 3a − 2b.
- A14i + 7j
- B−2i + 7j
- C−2i − 11j
- D−2i + 11j✓
Answer: 3a = 6i + 9j and 2b = 8i − 2j, so 3a − 2b = (6 − 8)i + (9 − (−2))j = −2i + 11j. (Mishandling the sign on −2b's j-term gives the +7j distractor.)
The points A and B have position vectors a = 3i − 2j and b = 7i + 5j. Find the vector AB.
- A−4i − 7j
- B4i + 7j✓
- C4i + 3j
- D10i + 3j
Answer: AB = b − a = (7 − 3)i + (5 − (−2))j = 4i + 7j. Working out a − b instead reverses the direction and gives −4i − 7j, and missing the double negative in 5 − (−2) gives 4i + 3j.
Which of the following vectors is parallel to 3i − 4j?
- A3i + 4j
- B6i − 8j✓
- C4i − 3j
- D6i + 8j
Answer: A parallel vector is a scalar multiple: 6i − 8j = 2(3i − 4j). The others are not scalar multiples (e.g. 4i − 3j swaps the components).
The points A and B have position vectors 2i + 5j and 8i − 3j. Find the position vector of the midpoint M of AB.
- A10i + 2j
- B3i + 4j
- C5i + j✓
- D6i − 4j
Answer: The midpoint is (a + b)/2 = ((2 + 8)/2)i + ((5 + (−3))/2)j = 5i + j. (The value 10i + 2j is the sum a + b without halving; the other values do not follow from the midpoint formula.)
Find the magnitude of the vector 2i − 3j + 6k.
- A11
- B√13
- C49
- D7✓
Answer: The magnitude is √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7. Adding the components gives 2 + 3 + 6 = 11; ignoring the k-term gives √(4 + 9) = √13; and omitting the square root leaves 49.
The points A and B have coordinates A(2, 1) and B(8, 13). The point P lies on AB such that AP : PB = 1 : 2. Find the coordinates of P.
- A(10, 14)
- B(4, 5)✓
- C(5, 7)
- D(6, 9)
Answer: AB = (8 − 2)i + (13 − 1)j = 6i + 12j. Since AP : PB = 1 : 2, AP = (1/3)AB = 2i + 4j, so P = A + AP = (2 + 2, 1 + 4) = (4, 5). The point (5, 7) is the midpoint of AB, and (6, 9) comes from reversing the ratio to 2 : 1.
Find the magnitude of the vector 3i + 4j.
- A25
- B7
- C5✓
- D3/4
Answer: |v| = √(3² + 4²) = √25 = 5. The value before square-rooting is the answer most often given, because that final step is the one left out.
Find the magnitude of the vector 5i + 12j.
- A13✓
- B17
- C169
- D5/12
Answer: |v| = √(5² + 12²) = √169 = 13. The value before square-rooting is the answer most often given, because that final step is the one left out.
Find the magnitude of the vector 8i + 6j.
- A100
- B14
- C10✓
- D4/3
Answer: |v| = √(8² + 6²) = √100 = 10. The value before square-rooting is the answer most often given, because that final step is the one left out.
Find the scalar product of (3i + 4j) and (2i − j).
- A8
- B6
- C5
- D2✓
Answer: Multiply matching components and add: (3)(2) + (4)(−1) = 6 − 4 = 2. A scalar product returns a NUMBER, not a vector — that is what 'scalar' is telling you.
Find the scalar product of (i + 2j) and (4i + 3j).
- A4
- B10✓
- C11
- D6
Answer: Multiply matching components and add: (1)(4) + (2)(3) = 4 + 6 = 10. A scalar product returns a NUMBER, not a vector — that is what 'scalar' is telling you.
Find the scalar product of (2i + 5j) and (3i + j).
- A11✓
- B6
- C17
- D5
Answer: Multiply matching components and add: (2)(3) + (5)(1) = 6 + 5 = 11. A scalar product returns a NUMBER, not a vector — that is what 'scalar' is telling you.
Points A and B have position vectors 2i + 3j and 7i − 9j. Find the vector AB.
- A5i − 12j✓
- B9i − 6j
- C−5i + 12j
- D5i + 12j
Answer: The vector from A to B is b − a = (7 − 2)i + (−9 − 3)j = 5i − 12j. Adding the position vectors instead gives 9i − 6j, and working out a − b reverses the direction to −5i + 12j.
Find the unit vector in the direction of 9i + 12j.
- A(3/7)i + (4/7)j
- B(9)i + (12)j
- C(3/5)i + (4/5)j✓
- D(1/25)i + (4/75)j
Answer: Divide the vector by its magnitude: |v| = √(9² + 12²) = 15, so the unit vector is (9/15)i + (12/15)j. Dividing by the sum of components, or by |v|², are the two usual errors — a unit vector must have magnitude exactly 1.
Vectors a = 2i + 3j and b = 6i + 9j. What is the relationship between them?
- AThey are equal in magnitude
- BThey are perpendicular, since a·b = 0
- Cb is parallel to a, since b = 3a✓
- DThey are neither parallel nor perpendicular
Answer: Check whether one is a scalar multiple of the other: 3(2i + 3j) = 6i + 9j = b, so they are parallel and point the same way. Parallel vectors are scalar multiples; perpendicular ones have a scalar product of zero.
What does it mean if two vectors have a scalar product of zero?
- AAt least one of the two vectors must be the zero vector
- BThey are parallel and point in exactly the same direction
- CThey have exactly the same magnitude as one another
- DThey are perpendicular✓
Answer: a·b = |a||b|cosθ, so a zero result with non-zero vectors forces cosθ = 0 and θ = 90°. It is the standard test for perpendicularity, and much quicker than comparing gradients.