S6 Maths: Vectors and the Dot Product: Paper 4 (Error clinic)
Ten items on the usual slips: treating a dot product as a vector, negative lengths, sign handling, confusing parallel with perpendicular, and dividing by a sum of components.
| 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 | Error clinic |
| # | Answer | One-line note |
|---|---|---|
| 1 | 23, a scalar | 2 × 4 + 3 × 5 = 23; the dot product multiplies matching components and adds, leaving a number. |
| 2 | 10 | √(36 + 64) = 10; simply adding the components to get 2 is a common slip. |
| 3 | 11 | 21 − 10 = 11; writing −2 × 5 as +10 wrongly gives 31. |
| 4 | Yes, since a · b = 0 | 36 − 36 = 0 makes them perpendicular; their lengths of 5 and 15 play no part. |
| 5 | Parallel, with a · b = 26 | b = 2a makes them parallel; the dot product 26 is not zero, so they are not perpendicular. |
| 6 | √2 ≈ 1.414 against 2 | The sum has length √2 ≈ 1.4142 while the lengths add to 2; equality needs the same direction. |
| 7 | k = −1 | 8k + 8 = 0 gives k = −1, and −8 + 8 = 0 checks it. |
| 8 | 180° | With cos θ = −10 ÷ 10 = −1 the angle is 180°, so the vectors point in opposite directions. |
| 9 | about 108.4° | a · b = −1 gives cos θ ≈ −0.3162, so θ ≈ 108.4°; the negative sign marks an obtuse angle. |
| 10 | The divisor is the length 5; the answer is (0.6, 0.8) | Divide by the length 5, not by 3 + 4, to get (0.6, 0.8). |
Work out the dot product and both lengths first, then substitute.Tutor Registration →
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