IB DP Mathematics: Analysis and Approaches – Vectors, lines and planes (HL) Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the vectors, lines and planes unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to syllabus sections 3.12–3.18 of the IB Mathematics: analysis and approaches guide, first assessment 2021, which stays in force until the new course is first assessed in May 2029. It applies to the May and November 2026, 2027 and 2028 HL sessions. Every question is HL only (AHL).

HL Paper 1 allows no technology; Papers 2 and 3 require it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give exact answers without a calculator, otherwise 3 significant figures, with angles in degrees. Vectors are written in a row, such as (2, −1, 3).

Learn the methods first in the vectors study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) The points A and B have coordinates A(3, −1, 2) and B(−1, 2, 14). Find the vector AB, the distance AB, and a unit vector in the direction of AB. [3]

2. (calculator-free) Let a = (2, −1, k) and b = (k, 4, −3), where k ∈ ℝ.

(a) Find the value of k for which a and b are perpendicular. [2] (b) For this value of k, find a × b and verify that |a × b| = |a||b|. [2]

3. (calculator-free) OABC is a rhombus, with OA = a and OC = c. Use the scalar product to prove that the diagonals OB and AC are perpendicular. [4]

4. (calculator allowed) Find the angle between the vectors u = (2, −3, 6) and v = (1, 4, −8). [3]

5. (calculator-free) The line L passes through A(2, 0, −1) and B(4, −3, 5).

(a) Find a vector equation of L. [2] (b) Write the equation of L in Cartesian form. [1] (c) Show that the point C(8, −9, 17) lies on L. [2]

6. (calculator allowed) Two drones fly at constant velocity. Their positions t seconds after take-off are

r_A = (1, 2, 0) + t(4, −1, 2) and r_B = (3, −3, 7) + t(2, 1, −1),

with distances in metres.

(a) Find the speed of drone A. [2] (b) Show that the flight paths intersect, and find the point of intersection. [3] (c) Determine whether the drones collide. Justify your answer. [1]

7. (calculator-free) The lines L₁ and L₂ have equations

L₁: r = (1, 2, 3) + λ(1, −1, 2) and L₂: r = (−2, −1, 4) + μ(2, 1, −1).

Determine whether L₁ and L₂ are parallel, intersecting or skew. [5]

8. (calculator-free) The points P(1, 2, 0), Q(3, 0, 1) and R(0, 1, 3) lie in the plane Π.

(a) Find PQ × PR. [2] (b) Find the exact area of triangle PQR. [2] (c) Find the Cartesian equation of Π. [2]

9. (calculator allowed) The line L has equation r = (2, −1, 3) + t(1, 2, −2) and the plane Π has equation 3x − y + 2z = 7.

(a) Find the coordinates of the point where L meets Π. [3] (b) Find the acute angle between L and Π. [2]

10. (calculator allowed) The planes Π₁ and Π₂ have equations x + 2y − z = 4 and 2x − y + z = 3.

(a) Find a vector equation of the line of intersection of Π₁ and Π₂. [4] (b) Find the acute angle between Π₁ and Π₂. [2]

11. (calculator-free) Consider the system of equations

x + y + z = 2 x − y + 2z = 1 3x + y + kz = m

where k, m ∈ ℝ.

(a) Find the value of k for which the system does not have a unique solution. [3] (b) For this value of k, find the value of m for which the system has infinitely many solutions. [2] (c) For these values of k and m, find a vector equation of the line of solutions. [2] (d) For this value of k and any other value of m, describe the geometric arrangement of the three planes. [1]

12. (calculator allowed) The plane Π has equation 2x − y + 2z = 8. The point A has coordinates (1, −1, −2).

(a) Write down a vector equation of the line through A perpendicular to Π. [1] (b) Find the coordinates of F, the point where this line meets Π. [3] (c) Hence find the distance AF. [1] (d) Find the coordinates of the reflection of A in Π. [1] (e) B(2, 2, 3) lies in Π. Find the angle between line AB and Π. [2]

Answers

1. AB = b − a = (−4, 3, 12) [1]. |AB| = √(16 + 9 + 144) = √169 = 13 [1]. Unit vector = (−4/13, 3/13, 12/13) [1] Examiner insight: A sign error in AB can still earn follow-through marks for the magnitude and unit vector, but only if the working is shown.

2. (a) a · b = 2k − 4 − 3k = −k − 4 [1]. Setting a · b = 0 gives k = −4 [1] (b) a = (2, −1, −4), b = (−4, 4, −3). a × b = ((−1)(−3) − (−4)(4), (−4)(−4) − (2)(−3), (2)(4) − (−1)(−4)) = (19, 22, 4) [1]. |a × b| = √(361 + 484 + 16) = √861, and |a||b| = √21 × √41 = √861, so |a × b| = |a||b| as sin 90° = 1 [1] Examiner insight: “Verify” needs both sides calculated and shown equal; quoting sin 90° = 1 alone does not earn the mark.

3. OB = a + c and AC = c − a [1]. OB · AC = (a + c) · (c − a) = c · c − a · a + a · c − c · a [1] = |c|² − |a|² since a · c = c · a [1]. A rhombus has equal sides, so |a| = |c| and OB · AC = 0; the diagonals are non-zero, so OB and AC are perpendicular [1] Examiner insight: The final mark needs the reason |a| = |c| and a conclusion; stopping at |c|² − |a|² loses it.

4. u · v = 2 − 12 − 48 = −58 [1]. |u| = 7 and |v| = 9, so cos θ = −58/63 [1], giving θ = 157° (157.0°) [1] Examiner insight: This is the angle between vectors, not lines, so the obtuse angle is correct; giving 23.0° loses the accuracy mark.

5. (a) Direction AB = (2, −3, 6) [1]. r = (2, 0, −1) + λ(2, −3, 6) [1] (b) (x − 2)/2 = y/(−3) = (z + 1)/6 [1] (c) Set (2, 0, −1) + λ(2, −3, 6) = (8, −9, 17): the x-component gives λ = 3 [1]. Then y = 0 − 9 = −9 and z = −1 + 18 = 17, so C lies on L [1] Examiner insight: The second mark needs the same λ checked in all three components, not just found from one.

6. (a) Speed = |(4, −1, 2)| = √(16 + 1 + 4) [1] = √21 = 4.58 m s⁻¹ [1] (b) With λ and μ: 1 + 4λ = 3 + 2μ, 2 − λ = −3 + μ, 2λ = 7 − μ [1]. From the second, μ = 5 − λ; the third gives 2λ = 2 + λ, so λ = 2 and μ = 3 [1]. Check x: 1 + 8 = 9 and 3 + 6 = 9, so the paths intersect at (9, 0, 4) [1] (c) A reaches (9, 0, 4) at t = 2 but B reaches it at t = 3, so the drones do not collide [1] Examiner insight: Using t for both drones in (b) tests for a collision, not crossing paths, and loses the method mark.

7. The directions (1, −1, 2) and (2, 1, −1) are not scalar multiples, so the lines are not parallel [1]. Equate: 1 + λ = −2 + 2μ, 2 − λ = −1 + μ, 3 + 2λ = 4 − μ [1]. Adding the first two: 3 = −3 + 3μ, so μ = 2 [1] and λ = 1 [1]. Third component: 3 + 2 = 5 but 4 − 2 = 2, and 5 ≠ 2, so the lines do not intersect: L₁ and L₂ are skew [1] Examiner insight: The first mark is for showing the directions are not parallel; without it, “no intersection” does not justify “skew”.

8. (a) PQ = (2, −2, 1) and PR = (−1, −1, 3) [1]. PQ × PR = ((−2)(3) − (1)(−1), (1)(−1) − (2)(3), (2)(−1) − (−2)(−1)) = (−5, −7, −4) [1] (b) |PQ × PR| = √(25 + 49 + 16) = √90 [1]. Area = ½√90 = 3√10/2 [1] (c) Use n = (5, 7, 4); d = n · p = 5 + 14 + 0 = 19 [1], so 5x + 7y + 4z = 19 [1] Examiner insight: Without a calculator, 4.74 is not exact; the accuracy mark needs 3√10/2 or an equivalent surd.

9. (a) Substitute x = 2 + t, y = −1 + 2t, z = 3 − 2t: 3(2 + t) − (−1 + 2t) + 2(3 − 2t) = 7 [1]. This simplifies to 13 − 3t = 7, so t = 2 [1]. The point is (4, 3, −1) [1] (b) sin φ = |(1, 2, −2) · (3, −1, 2)|/(3 × √14) = 3/(3√14) = 1/√14 [1], so φ = 15.5° [1] Examiner insight: Using cos gives 74.5°, the angle with the normal; the method mark needs sin φ or 90° minus that angle.

10. (a) Direction = n₁ × n₂ = (1, 2, −1) × (2, −1, 1) [1] = (1, −3, −5) [1]. Put z = 0: x + 2y = 4 and 2x − y = 3 give x = 2, y = 1 [1]. r = (2, 1, 0) + t(1, −3, −5) [1] (b) cos θ = |n₁ · n₂|/(|n₁||n₂|) = |2 − 2 − 1|/(√6 × √6) = 1/6 [1], so θ = 80.4° [1] Examiner insight: Without the modulus the normals give 99.6°, which loses the accuracy mark for an acute angle.

11. (a) Eliminate x: (first) − (second) gives 2y − z = 1, and (third) − 3 × (first) gives −2y + (k − 3)z = m − 6 [1]. Adding: (k − 4)z = m − 5 [1]. z cannot be found uniquely when k − 4 = 0, so k = 4 [1] (b) With k = 4, 2 × (first) + (second) gives 3x + y + 4z = 5 [1]. The third equation matches this only when m = 5 [1] (c) Adding the first two equations: 2x + 3z = 3. Let z = 2s: x = 3/2 − 3s and y = 1/2 + s [1]. r = (3/2, 1/2, 0) + s(−3, 1, 2) [1] (d) No two normals are parallel, so the planes meet in pairs in three parallel lines: a triangular prism with no common point [1] Examiner insight: In (d), “no solution” alone does not earn the mark; you must name the triangular prism arrangement.

12. (a) r = (1, −1, −2) + t(2, −1, 2) [1] (b) Substitute: 2(1 + 2t) − (−1 − t) + 2(−2 + 2t) = 8 [1]. This gives 9t − 1 = 8, so t = 1 [1]. F = (3, −2, 0) [1] (c) AF = |(2, −1, 2)| = 3 [1] (d) The reflection is at t = 2: (5, −3, 2) [1] (e) AB = (1, 3, 5), and sin φ = |(1, 3, 5) · (2, −1, 2)|/(√35 × 3) = 9/(3√35) [1], so φ = 30.5° [1] Examiner insight: “Hence” in (c) means the distance must come from (b); an unrelated method may not earn the mark.

Where marks are usually lost

• One parameter for both lines, as in questions 6 and 7, instead of λ and μ.
• Checking only two components for an intersection or a point on a line.
• “Skew” claimed without showing the directions are not parallel.
• cos instead of sin for a line–plane angle, as in questions 9 and 12.
• An obtuse angle between planes when the acute angle is asked for.
• Sign errors in the middle component of a vector product.
• The missing ½ for the area of a triangle.
• Decimals on calculator-free questions where a surd is expected.
• A system with no solutions left without its geometric reason.

Next steps

• Recap the key facts in the vectors revision notes.
• Re-learn any weak topic in the vectors, lines and planes study guide.
• See the rest of the course on the IB DP Maths AA course hub and tick off the printable syllabus checklist.
• Plan with the AA exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).