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S5 Maths: Vectors – Mixed Applications (Paper 3: Challenge)
S5 Maths: Vectors – Mixed Applications (Paper 3: Challenge)
Handle mixed problems with section points, angles and square expansion, with all values computed from components and trigonometry.
Direct answers This paper joins vectors with geometry and lists both exact and approximate values.
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: S5 Maths: Vectors |
| Level | Challenge |
02 Questions (10 items)
- FoundationGiven A(1, 2) and B(5, 10), P on AB with AP to PB as 3 to 1, find P.
- FoundationIn triangle ABC with A(0, 0), B(6, 0) and C(0, 8), find AB dot AC and the area.
- FoundationGiven a = (2, 1) and b = (1, -2), find a dot b and |a + b|.
- IntermediateGiven a = (1, 2) and b = (3, 1), find t so that a + t b is perpendicular to a.
- IntermediateGiven |a| = 3, |b| = 4 and an angle of 60 degrees, find a dot b and |a + b|.
- IntermediateGiven |a| = 2, |b| = 5 and a dot b = -8, find the angle between them.
- ChallengeGiven A(1, 1), B(5, 3) and C(3, 7), find angle BAC.
- ChallengeIn parallelogram ABCD with A(0, 0), B(4, 1) and D(1, 3), find C and the diagonal intersection.
- ChallengeFind a unit vector perpendicular to (3, 4).
- Error redoA student concludes b = c from a dot b = a dot c. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | (4, 8) | P covers three quarters of the segment, adding 3 across and 6 up. |
| 2 | AB dot AC = 0, area 24 | The two vectors lie on the axes so the dot product is zero, and the area is 6 times 8 over 2. |
| 3 | a dot b = 0, |a + b| about 3.16 (root 10) | The dot product is 2 minus 2, and a + b is (3, -1) with length about 3.16. |
| 4 | t = -1 | Expanding the dot product gives |a| squared plus t times a dot b, so 5 plus 5t is zero. |
| 5 | a dot b = 6, |a + b| about 6.08 (root 37) | The dot product is 3 times 4 times 0.5, and the squares add to 9 plus 16 plus 12, giving 37. |
| 6 | About 143.13 degrees | The cosine is negative 8 over 10, and the inverse cosine gives about 143.13 degrees. |
| 7 | 45 degrees | The dot product of AB and AC is 20 while the lengths are root 20 and root 40, giving a cosine near 0.7071. |
| 8 | C(5, 4), intersection (2.5, 2) | C equals B plus D, and the intersection is the midpoint of A and C. |
| 9 | (-0.8, 0.6) | Swap the components and change one sign to get (-4, 3), then divide by the length 5. |
| 10 | a cannot be cancelled; it only shows a is perpendicular to b – c | Rearranging gives a dot (b – c) equal to zero, which only says a is perpendicular to the difference. |
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.
Sketch the situation first, then decide between a dot product and a ratio.Tutor Registration →
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