IB DP Mathematics: Applications and Interpretation – Radians, the unit circle, matrix transformations and vectors (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 HL unit on radians, the unit circle, matrix transformations and vectors in IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 3.7–3.13, and every question is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

All three HL papers in this course require technology, so every question is labelled “(calculator allowed)”. Give answers exactly or to 3 significant figures. Use radians unless the question says degrees. Vectors are written in a row, such as (2, −1, 3).

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

Questions

1. (calculator allowed, HL only)

(a) Write 72° in radians as an exact multiple of π. [1] (b) Convert 2.5 radians to degrees. [2]

2. (calculator allowed, HL only) A sector of a circle has radius 12 cm and arc length 15 cm.

(a) Find the angle of the sector in radians. [2] (b) Find the area of the sector. [2] (c) Find the area of the segment between the arc and the chord. [2]

3. (calculator allowed, HL only) Given that sin θ = 5/13 and π/2 < θ < π, find the exact values of cos θ and tan θ. [4]

4. (calculator allowed, HL only) In triangle PQR, PQ = 11 cm, QR = 8 cm and angle QPR = 35°.

(a) Find the two possible values of angle PRQ, in degrees. [3] (b) Find PR when angle PRQ is obtuse. [3]

5. (calculator allowed, HL only) Solve 3 cos(2x) = x − 1 for 0 ≤ x ≤ π. [4]

6. (calculator allowed, HL only) Transformation T is a reflection in the line y = x followed by an enlargement with centre the origin and scale factor 3.

(a) Write the matrix for each transformation and hence the single matrix for T. [3] (b) Find the image of the point (2, −1) under T. [1] (c) A shape has area 4.5 units². Find the area of its image under T. [2]

7. (calculator allowed, HL only) A transformation maps (x, y) to (x’, y’) where

( x' )   ( 0.8  −0.6 ) ( x )   ( 2 )
( y' ) = ( 0.6   0.8 ) ( y ) + ( 1 )

(a) Describe fully the transformation given by the matrix, stating the angle in radians. [2] (b) Find the point whose image is (5, 4). [3]

8. (calculator allowed, HL only) A model aircraft starts at the point (0, 0, 5) and flies at a constant speed of 18 m s⁻¹ in the direction 2i − j + 2k. Distances are in metres.

(a) Find its velocity vector. [2] (b) Find its position after 4 seconds. [2]

9. (calculator allowed, HL only) Line L₁ has equation r = (1, 2, 0) + λ(2, −1, 3). Line L₂ has equation r = (2, 3, 3) + μ(1, −1, 1).

(a) Write L₁ in parametric form. [1] (b) Show that L₁ and L₂ intersect and find the point of intersection. [4] (c) Find the acute angle between L₁ and L₂, in degrees. [3]

10. (calculator allowed, HL only) Two boats move with constant velocity. Distances are in km, i points east and j points north, and t is the time in hours after 12:00. Boat A has position r_A = t(4, 3). Boat B has position r_B = (15, 5) + t(−1, −2).

(a) Find the speed of boat A. [1] (b) Find the position of B relative to A, AB, in terms of t. [2] (c) Show that |AB|² = 50t² − 200t + 250. [2] (d) Find the time when the boats are closest and the minimum distance between them. [3] (e) Find the time when B is due east of A, and the distance between them then. [2]

11. (calculator allowed, HL only) A ball is thrown from the point (0, 1.5). Its velocity t seconds later is v = (14, 12 − 9.8t) m s⁻¹, with distances in metres and the ground at y = 0.

(a) Find the position vector of the ball at time t. [3] (b) Find the time when the ball hits the ground. [2] (c) Find the horizontal distance travelled. [1] (d) Find the speed of the ball as it hits the ground. [2]

12. (calculator allowed, HL only) The points A(1, 0, 2), B(3, 1, 0) and C(0, 4, 3) are given.

(a) Show that angle BAC is a right angle. [2] (b) Find AB × AC. [2] (c) Hence find the area of triangle ABC. [2] (d) A force F = (5, 2, −1) N acts at A. Find the component of F in the direction of AB. [2] (e) Find the component of F perpendicular to AB, in the plane of F and AB. [2]

Answers

1. (a) 72 × π/180 = 2π/5 [1] (b) 2.5 × 180/π [1] = 143° (143.2…) [1] Examiner insight: in (b) a bare 143 with no conversion shown can lose the method mark if it is wrong; write the × 180/π step.

2. (a) l = rθ, so 15 = 12θ [1], θ = 1.25 rad [1] (b) A = ½ × 12² × 1.25 [1] = 90 cm² [1] (c) Triangle area = ½ × 12² × sin 1.25 = 68.33… [1]; segment = 90 − 68.33 = 21.7 cm² [1] Examiner insight: sin 1.25 must be evaluated in radian mode; a degree-mode value gives 88.4 cm² and loses the accuracy mark.

3. cos²θ = 1 − 25/169 = 144/169 [1] θ in the second quadrant, so cos θ < 0: cos θ = −12/13 [1] tan θ = sin θ / cos θ = (5/13)/(−12/13) [1] = −5/12 [1] Examiner insight: +12/13 earns the method mark but not the accuracy mark; state why the sign is negative.

4. (a) sin R / 11 = sin 35° / 8, so sin R = 11 sin 35° / 8 = 0.7887 [1] R = 52.1° [1] or R = 180° − 52.1° = 127.9° [1] (b) Angle PQR = 180° − 35° − 127.9° = 17.1° [1] PR / sin 17.06° = 8 / sin 35° [1], PR = 4.09 cm [1] Examiner insight: carry the unrounded angle (127.938…) into (b); rounding to 127.9° first gives 4.10 cm and loses the accuracy mark.

5. Graph y = 3 cos(2x) and y = x − 1 (or y = 3 cos(2x) − x + 1) [1] on 0 ≤ x ≤ π in radian mode [1] x = 0.816 [1], x = 2.65 [1] Examiner insight: a sketch or the equation you graphed secures the method marks; listing roots outside 0 ≤ x ≤ π loses an accuracy mark.

6. (a) Reflection R = (0 1; 1 0) [1], enlargement E = (3 0; 0 3) [1] T = ER = (0 3; 3 0) [1] (b) T(2, −1) = (0×2 + 3×(−1), 3×2 + 0×(−1)) = (−3, 6) [1] (c) det T = 0 − 9 = −9, so area scale factor |−9| = 9 [1]; image area = 9 × 4.5 = 40.5 units² [1] Examiner insight: the image area uses |det T| = 9; an answer of −40.5 units² loses the accuracy mark in (c).

7. (a) Rotation about the origin, anticlockwise [1], through θ where cos θ = 0.8, sin θ = 0.6, so θ = 0.644 rad [1] (b) A(x, y) + (2, 1) = (5, 4), so A(x, y) = (3, 3) [1] det A = 1, A⁻¹ = (0.8 0.6; −0.6 0.8) [1] (x, y) = (0.8×3 + 0.6×3, −0.6×3 + 0.8×3) = (4.2, 0.6) [1] Examiner insight: subtract the translation before applying A⁻¹; inverting first and then subtracting is a method error.

8. (a) |2i − j + 2k| = √9 = 3 [1]; v = 18 × (2i − j + 2k)/3 = 12i − 6j + 12k m s⁻¹ [1] (b) r = (0, 0, 5) + 4(12, −6, 12) [1] = (48, −24, 53) [1] Examiner insight: using the direction vector as the velocity (speed 3, not 18) scores nothing in (a) and loses the accuracy mark in (b).

9. (a) x = 1 + 2λ, y = 2 − λ, z = 3λ [1] (b) 1 + 2λ = 2 + μ and 2 − λ = 3 − μ [1] Solving: λ = 2, μ = 3 [1] z: 3λ = 6 and 3 + μ = 6, consistent [1] Point of intersection (5, 0, 6) [1] (c) (2, −1, 3) · (1, −1, 1) = 6 [1] cos θ = 6 / (√14 × √3) [1], θ = 22.2° [1] Examiner insight: “show that they intersect” needs the third equation checked in writing; solving two equations alone does not prove it.

10. (a) |(4, 3)| = 5 km h⁻¹ [1] (b) AB = r_B − r_A [1] = (15 − 5t, 5 − 5t) [1] (c) |AB|² = (15 − 5t)² + (5 − 5t)² [1] = 225 − 150t + 25t² + 25 − 50t + 25t² = 50t² − 200t + 250 [1] (d) d/dt(50t² − 200t + 250) = 100t − 200 = 0 [1], t = 2, so 14:00 [1] |AB| = √(50×4 − 400 + 250) = √50 = 7.07 km [1] (e) Same y-coordinate: 3t = 5 − 2t, so t = 1 (13:00) [1] A is at (4, 3) and B at (14, 3): distance 10 km [1] Examiner insight: in (c) every expanded line must be shown, since the answer is given; in (d) giving t = 2 but not the distance loses the final mark.

11. (a) Integrate: r = (14t + c₁, 12t − 4.9t² + c₂) [1] At t = 0, r = (0, 1.5), so c₁ = 0, c₂ = 1.5 [1] r = (14t, 1.5 + 12t − 4.9t²) [1] (b) 1.5 + 12t − 4.9t² = 0 [1], t = 2.57 s (positive root) [1] (c) x = 14 × 2.568… = 36.0 m [1] (d) v = (14, 12 − 9.8 × 2.568…) = (14, −13.17) [1]; speed = √(14² + 13.17²) = 19.2 m s⁻¹ [1] Examiner insight: omitting the constant 1.5 in (a) loses the accuracy mark there, and later parts only earn follow-through marks.

12. (a) AB = (2, 1, −2), AC = (−1, 4, 1) [1] AB · AC = −2 + 4 − 2 = 0, so angle BAC = 90° [1] (b) AB × AC = (1×1 − (−2)×4, (−2)(−1) − 2×1, 2×4 − 1×(−1)) [1] = (9, 0, 9) [1] (c) |AB × AC| = √162 = 9√2 [1]; area = ½ × 9√2 = 6.36 units² [1] (d) F · AB / |AB| = (10 + 2 + 2)/3 [1] = 4.67 N [1] (e) F × AB = (−3, 8, 1), |F × AB| = √74 [1]; √74 / 3 = 2.87 N [1] Examiner insight: “hence” in (c) means use the vector product from (b); a method that ignores it may not earn full marks.

Where marks are usually lost

• Degree-mode trig in a radian question, such as sin 1.25 or cos(2x).
• Finding only one angle in an ambiguous-case sine rule question that asks for “two possible values”.
• Graphical solutions given without a sketch or the graphed equations, so no method mark can be awarded.
• Composing matrices in the order written in the question, instead of first-on-the-right.
• Forgetting the modulus of a negative determinant when finding an image area.
• Treating a direction vector as the velocity without rescaling to the given speed.
• Not checking the third equation when showing two 3D lines intersect.
• Premature rounding in multi-step kinematics, which shifts the final 3 s.f. answer.
• Dropping units (km, m s⁻¹, N) on final answers in context.

Next steps

• Revisit the revision notes for the formula and matrix tables.
• Rework any weak section from the study guide.
• See the whole course on the IB DP Maths AI course hub and tick off topics on the printable syllabus checklist.
• Read the AI exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021 — syllabus sections 3.7–3.13 (AHL).