OxfordAQA International AS and A-level Mathematics: Scalar product and perpendicular distance (9660) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section P2.10 (second part): the scalar product, the angle between two lines, and the foot of the perpendicular and perpendicular distance from a point to a line. The content belongs to International A-level Unit P2 and is not examined at International AS. Vectors are written across the page as (x, y, z); in a paper they may appear as columns. Give exact answers unless a question asks for a decimal.

Study first with the study guide or the revision notes. The course hub is OxfordAQA A-level Mathematics, the checklist tracks your progress, and the free diagnostics show where to start.

Questions

1. The vectors a = (3, −2, 5) and b = (−1, 4, 2) are given. Find a.b and state, with a reason, whether the angle between a and b is acute or obtuse. [2]

2. The vectors u = (k, −5, −2) and v = (k, k, 3) are perpendicular. Find the possible values of k. [4]

3. Two lines have equations

r = (−2, 5, 1) + λ(3, −4, 12) and r = (4, 0, −3) + μ(2, 3, 6).

Find the acute angle between the lines, correct to 1 decimal place. [4]

4. The points A, B and C have coordinates A(0, 1, 3), B(1, −1, 5) and C(2, 4, 5).

(a) Show that angle BAC is a right angle. [2] (b) Find the exact area of triangle ABC. [3]

5. The point P has coordinates (2, 0, 7). The line l has equation r = (1, 0, 2) + t(1, 2, −2).

(a) Find the coordinates of the foot of the perpendicular from P to l. [4] (b) Hence find the perpendicular distance from P to l. [2]

6. A line in two dimensions has equation r = (−1, 5) + t(3, −2). Find the perpendicular distance from the point (4, 6) to this line. [4]

7. The line l₁ has direction (3, 0, 4) and the line l₂ has direction (2, p, 1), where p is a constant. The acute angle θ between the lines satisfies cos θ = 2/3. Find the possible values of p. [5]

8. A drone flies along the line r = (−20, 10, 2) + t(12, −4, 3). Distances are in hundreds of metres, t is the time in minutes and the axes point east, north and upwards. A sensor stands at the point P(5, 5, 8).

(a) Find the value of t when the drone is closest to the sensor. [4] (b) Find the least distance between the drone and the sensor, in metres to 3 significant figures. [2] (c) Find the speed of the drone in metres per minute. [1]

9. Show that the acute angle between the lines y = 2x − 1 and y = −3x + 4 is 45°. [4]

10. A straight cable has direction vector (2, −3, 6). Find the angle between the cable and the upward vertical direction k = (0, 0, 1), to 1 decimal place. [3]

11. The points A, B and C have coordinates A(7, 1, 3), B(1, −2, 3) and C(7, −5, 9).

(a) Find angle ABC, correct to 1 decimal place. [4] (b) Find the coordinates of the point D, the foot of the perpendicular from A to the line BC. [4] (c) Hence find the area of triangle ABC. [2]

Answers

1. a.b = (3)(−1) + (−2)(4) + (5)(2) = −3 − 8 + 10 = −1 [1] The scalar product is negative, so cos θ is negative and the angle is obtuse [1] [2] Examiner insight: A reason is usually needed with the word “obtuse”. Link the sign of a.b to the sign of cos θ.

2. u.v = k² − 5k − 6 [1] Set u.v = 0 for perpendicular vectors: k² − 5k − 6 = 0 [1] (k − 6)(k + 1) = 0 [1] k = 6 or k = −1 [1] [4] Examiner insight: Perpendicular means u.v = 0 and no magnitudes are needed. Give both roots, because a quadratic condition has two solutions.

3. Direction vectors (3, −4, 12) and (2, 3, 6); the position vectors are not used. d₁.d₂ = 6 − 12 + 72 = 66 [1] |d₁| = √(9 + 16 + 144) = 13 and |d₂| = √(4 + 9 + 36) = 7 [1] cos θ = 66 / (13 × 7) = 66/91 [1] θ = 43.5° [1] [4] Examiner insight: Keep the fraction 66/91 until the last step. The scalar product is positive here, so no supplement is needed.

4. (a) AB = (1, −2, 2) and AC = (2, 3, 2) [1] AB.AC = 2 − 6 + 4 = 0, so AB is perpendicular to AC and angle BAC = 90° [1] [2] (b) |AB| = √(1 + 4 + 4) = 3 [1] |AC| = √(4 + 9 + 4) = √17 [1] Area = ½ × 3 × √17 = 3√17/2 [1] [3] Examiner insight: In a “show that” question, write the vectors, the calculation and the conclusion. Part (b) uses the right angle from part (a), so state that it is the base and height.

5. (a) X = (1 + t, 2t, 2 − 2t) [1] PX = X − P = (t − 1, 2t, −5 − 2t); PX.d = (t − 1) + 4t − 2(−5 − 2t) = 9t + 9 [1] 9t + 9 = 0, so t = −1 [1] Foot: (0, −2, 4) [1] [4] (b) PX = (−2, −2, −3) (or its negative) [1] Distance = √(4 + 4 + 9) = √17 [1] [2] Examiner insight: Substitute t back to get a point. Checking PX.d = 0 with your numbers is a quick way to find sign slips.

6. X = (−1 + 3t, 5 − 2t), PX = X − P = (3t − 5, −1 − 2t) [1] PX.d = 3(3t − 5) − 2(−1 − 2t) = 13t − 13 = 0 [1] t = 1, so X = (2, 3) and PX = (−2, −3) [1] Distance = √13 [1] [4] Examiner insight: The same method works in two dimensions. Give the distance as a surd unless a decimal is requested.

7. d₁.d₂ = 6 + 0 + 4 = 10 for every p [1] |d₁| = 5 and |d₂| = √(5 + p²) [1] 10 / (5√(5 + p²)) = 2/3 [1] √(5 + p²) = 3, so 5 + p² = 9 [1] p = 2 or p = −2 [1] [5] Examiner insight: Notice that the scalar product does not depend on p. Both roots remain valid, so state both.

8. (a) X = (−20 + 12t, 10 − 4t, 2 + 3t) [1] X − P = (12t − 25, 5 − 4t, 3t − 6) [1] (X − P).d = 12(12t − 25) − 4(5 − 4t) + 3(3t − 6) = 169t − 338 = 0 [1] t = 2 minutes [1] [4] (b) At t = 2, X = (4, 2, 8) and X − P = (−1, −3, 0) [1] Distance = √10 hundred metres = 316 m (3 s.f.) [1] [2] (c) |d| = √(144 + 16 + 9) = 13 hundred metres per minute = 1300 m per minute [1] [1] Examiner insight: Read the units in the stem. The distance in the units of the question is √10, and the final answer has to be converted to metres.

9. Direction vectors (1, 2) and (1, −3) [1] d₁.d₂ = 1 − 6 = −5, |d₁| = √5, |d₂| = √10 [1] cos θ = −5 / √50 = −1/√2, so the vectors are at 135° [1] The acute angle between the lines is 180° − 135° = 45° [1] [4] Examiner insight: A negative cosine gives an obtuse angle. A “show that” question wants the step from 135° to 45° written out.

10. d.k = 6 and |d| = √(4 + 9 + 36) = 7 [1] cos θ = 6/7 [1] θ = 31.0° [1] [3] Examiner insight: k is already a unit vector, so |k| = 1. Showing cos θ = 6/7 first gives a clear line of working.

11. (a) BA = (6, 3, 0) and BC = (6, −3, 6) [1] BA.BC = 36 − 9 + 0 = 27, |BA| = 3√5, |BC| = 9 [1] cos ABC = 27 / (27√5) = 1/√5 [1] angle ABC = 63.4° [1] [4] (b) Line BC: r = (1, −2, 3) + t(6, −3, 6) [1] X − A = (6t − 6, −3t − 3, 6t), and (X − A).(6, −3, 6) = 81t − 27 = 0 [1] t = 1/3 [1] D = (3, −3, 5) [1] [4] (c) AD = |(−4, −4, 2)| = 6 [1] Area = ½ × BC × AD = ½ × 9 × 6 = 27 [1] [2] Examiner insight: Both vectors in part (a) must start at B. Part (c) is a “hence” question, so use AD from part (b) instead of starting again.

Where marks are usually lost

• Using position vectors, not direction vectors, when finding the angle between two lines.
• Dividing by |u| only, or forgetting the denominator in cos θ.
• Sign slips with negative components. Bracket every product.
• Giving an obtuse angle when asked for the angle between lines, or the reverse.
• Solving PX.d = 0 and stopping at t. The foot is a point.
• Using the distance from P to a point on the line, not the foot.
• Early rounding that carries into the final decimal.
• Skipping lines in a “show that” argument.
• Losing track of units, such as hundreds of metres.

Next steps

• Re-read the study guide for any question you missed.
• Condense your learning with the revision notes.
• Visit the course hub and tick items off the checklist.
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Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section P2.10 (Vectors), published by OxfordAQA. For International AS exams from May/June 2018 and International A-level exams from May/June 2019.