A Level Maths: Pure 3 chapter 7: Vectors MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

26 exam-style questions on Vectors, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

26 questions · 1 topics · 10 per test

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What chapter 7 covers

Ten Vectors questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. Find the magnitude of the vector 2i − 3j + 6k.

  1. A
    7
  2. B
    5
  3. C
    49
  4. D
    √31
Show answer

Answer: A. |2i − 3j + 6k| = √(22 + (−3)2 + 62) = √49 = 7. (−3)2 is +9, not −9 (which would give √31), and adding the components (to get 5) does not give the length.

2. Find the acute angle between the lines r = 2i + j + s(3i − 4k) and r = i − j + 3k + t(i + 2j + 2k).

  1. A
    109.5°
  2. B
    70.5°
  3. C
    82.3°
  4. D
    19.5°
Show answer

Answer: B. Use the direction vectors: (3i − 4k)·(i + 2j + 2k) = 3 − 8 = −5, with magnitudes 5 and 3, so cos θ = −1/3 and θ = 109.5°. The acute angle is 180° − 109.5° = 70.5°. 82.3° comes from using the position vectors instead of the direction vectors.

3. Find a vector equation of the line through A(2, −1, 3) and B(5, 1, −1).

  1. A
    r = 3i + 2j − 4k + t(2i − j + 3k)
  2. B
    r = 2i − j + 3k + t(7i + 2k)
  3. C
    r = 2i − j + 3k + t(3i + 2j − 4k)
  4. D
    r = 5i + j − k + t(2i − j + 3k)
Show answer

Answer: C. Use the position vector of a point on the line, OA = 2i − j + 3k, and the direction AB = OB − OA = 3i + 2j − 4k. Swapping the roles of the point and the direction gives a different line; 7i + 2k is OA + OB, which is not along the line.

4. The points A, B and C have coordinates (1, 2, 0), (3, 0, 1) and (2, 4, 3). Find angle BAC.

  1. A
    95.1°
  2. B
    58.1°
  3. C
    54.4°
  4. D
    84.9°
Show answer

Answer: D. Use vectors from A: AB = 2i − 2j + k and AC = i + 2j + 3k. AB·AC = 2 − 4 + 3 = 1, |AB| = 3 and |AC| = √14, so cos BAC = 1/(3√14) and angle BAC = 84.9°. Using BA with AC gives 95.1°, using the position vectors OB and OC gives 58.1°, and 54.4° is angle ABC.

5. In the vector equation of a line, r = a + tb, what does b represent?

  1. A
    the position vector of a point on the line
  2. B
    a vector in the direction of the line
  3. C
    a vector perpendicular to the line
  4. D
    the shortest distance from the origin
Show answer

Answer: B. a is the position vector of one point on the line and b is a direction vector: as t varies, r moves along the line in the direction of b.

6. Two lines in three dimensions are described as skew. What does this mean?

  1. A
    They are not parallel and do not intersect
  2. B
    They are parallel and do not intersect
  3. C
    They intersect at right angles
  4. D
    They lie in the same plane but never meet
Show answer

Answer: A. In three dimensions two lines that never meet need not be parallel. Skew lines are neither parallel nor intersecting, so no plane contains both of them.

7. Find the scalar product (2i − j + 3k)·(i + 4j − 2k).

  1. A
    −8
  2. B
    8
  3. C
    4
  4. D
    2i − 4j − 6k
Show answer

Answer: A. Multiply matching components and add: 2 × 1 + (−1) × 4 + 3 × (−2) = 2 − 4 − 6 = −8. The scalar product is a number, not a vector; 4 comes from a sign slip in the last product.

8. Given that |pi + 2j − 2k| = 5, find the possible values of p.

  1. A
    p = ±√17
  2. B
    p = √17 only
  3. C
    p = ±5
  4. D
    p = ±√33
Show answer

Answer: A. p2 + 4 + 4 = 25 gives p2 = 17, so p = ±√17: both signs give the same magnitude. ±5 comes from treating (−2)2 as −4, and ±√33 from p2 − 8 = 25.

9. N is the foot of the perpendicular from the point P, with position vector p, to the line r = a + tb. Which equation gives the value of t at N?

  1. A
    (a + tb)·b = 0
  2. B
    (a + tb − p)·p = 0
  3. C
    (a + tb)·p = 0
  4. D
    (a + tb − p)·b = 0
Show answer

Answer: D. PN = ON − OP = a + tb − p, and PN must be perpendicular to the direction of the line, so PN·b = 0. (a + tb)·b = 0 would make ON, not PN, perpendicular to the line.

10. Find the coordinates of the foot of the perpendicular from the point P(4, 3, 2) to the line r = i + 2k + t(2i + j − 2k).

  1. A
    (7, 3, −4)
  2. B
    (−1, −1, 4)
  3. C
    (3, 1, 0)
  4. D
    (−1, −2, −2)
Show answer

Answer: C. A general point of the line is N(1 + 2t, t, 2 − 2t), so PN = (2t − 3)i + (t − 3)j − 2tk. PN is perpendicular to 2i + j − 2k: 2(2t − 3) + (t − 3) + 4t = 0, so 9t = 9, t = 1 and N is (3, 1, 0). (−1, −2, −2) is the vector PN, not the point N.

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How many Vectors MCQs are there?

26, each with the answer and a short explanation, across 1 topics: Vectors. Every chapter test draws ten of them, so you can take it many times.

Is the A Level Maths: Pure 3 chapter 7 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

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