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S6 Maths: Vectors and the Dot Product: Paper 3 (Challenge)
S6 Maths: Vectors and the Dot Product: Paper 3 (Challenge)
This paper handles negative dot products and obtuse angles, triangle angles, projections, and expanding |a + b|.
Direct answers Items 1-3: about 98.1°, about 81.9°, then 4 and (4, 0). Items 4-6: about 4.3589, then 5 with 58, then 0. Items 7-9: −1 with 2.6458, −5, about 9.8995. Item 10: cos θ = 0.96.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Maths: Vectors and the Dot Product |
| Level | Challenge |
02 Questions (10 items)
- FoundationGiven a = (2, −1) and b = (1, 3), find the angle (1 d.p.).
- FoundationIn triangle ABC with A(1, 1), B(5, 4) and C(2, 7), find angle ABC (1 d.p.).
- FoundationGiven a = (4, 3) and b = (2, 0), find the scalar and vector projections of a on b.
- IntermediateGiven |a| = 2, |b| = 3 and a · b = 3, find |a + b| (4 d.p.).
- IntermediateGiven a = (2, k) is parallel to b = (4, 10), find k and a · b.
- IntermediateGiven a = (5, 2) and b = (2, 5), find (a + b) · (a − b) and describe the relation.
- ChallengeGiven |a| = 2 and |b| = 1 with an angle of 120°, find a · b and |a − b| (4 d.p.).
- ChallengeGiven a = (1, 2) and b = (3, −1), find t so that a + t b is perpendicular to a.
- ChallengeGiven a = (6, 8) and b = (1, 1), find the scalar projection of a on b (4 d.p.).
- Error redoA student uses cos θ = a · b ÷ (|a| + |b|) = 2.4 for a = (3, 4), b = (4, 3). What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | about 98.1° | a · b = −1 with lengths √5 and √10 gives cos θ ≈ −0.1414, so θ ≈ 98.1°. |
| 2 | about 81.9° | BA = (−4, −3) and BC = (−3, 3) give a dot product of 3, so cos θ ≈ 0.1414 and the angle is 81.9°. |
| 3 | scalar projection 4, vector projection (4, 0) | The scalar projection is 8 ÷ 2 = 4, and the vector projection is (4, 0). |
| 4 | about 4.3589 | |a + b|² = 4 + 9 + 6 = 19, so the length is √19 ≈ 4.3589. |
| 5 | k = 5, a · b = 58 | Parallel vectors have proportional components, so k = 5; then a · b = 8 + 50 = 58. |
| 6 | 0, so they are perpendicular | The product equals |a|² − |b|² = 0, matching the diagonals of a rhombus. |
| 7 | a · b = −1, |a − b| ≈ 2.6458 | The dot product is 2 × 1 × cos 120° = −1, so |a − b|² = 7 and the length is 2.6458. |
| 8 | t = −5 | (a + t b) · a = 5 + t(1) = 0, so t = −5. |
| 9 | about 9.8995 | With a · b = 14 and |b| = √2, the projection is 14 ÷ 1.4142 ≈ 9.8995. |
| 10 | The denominator should be |a||b|; cos θ = 0.96 and θ ≈ 16.3° | The correct value is 24 ÷ 25 = 0.96; a cosine can never exceed 1, which exposes the mistake. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Work out the dot product and both lengths first, then substitute.Tutor Registration →
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