June 2022 Paper 2 Q2
2 The points \(A\) and \(B\) have position vectors \(3\mathbf{i} + 2\mathbf{j}\) and \(4\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}\) respectively.
Point \(P\) has position vector \(p\mathbf{i} - 3\mathbf{k}\), where \(p\) is a constant. \(P\) lies on the circumference of a circle of which \(AB\) is a diameter.
| Scheme | Marks | AO |
|---|---|---|
| \((4\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}) - (3\mathbf{i} + 2\mathbf{j})\) \((= \mathbf{i} - 5\mathbf{k})\) or \((3\mathbf{i} + 2\mathbf{j}) - (4\mathbf{i} + 2\mathbf{j} - 5\mathbf{k})\) \((= 5\mathbf{k} - \mathbf{i})\) | M1 | 1.1 |
| \(AB = \sqrt{26}\) or 5.10 (3 sf) or 5.1 | A1 | 1.1 |
| [2] |
Notes
M1: \(\mathbf{b} - \mathbf{a}\) or \(\mathbf{a} - \mathbf{b}\) attempted, using \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) or column vectors
May be implied by calculation seen
A1: www. Correct answer, no working: M1A1
Mark(s) cannot be gained retrospectively in (b)
| Scheme | Marks | AO |
|---|---|---|
| ‘26’ \(= (p - 3)^2 + 4 + 9 + (p - 4)^2 + 4 + 4\) | M1 | 3.1a |
| \(p^2 - 7p + 10 = 0\) oe or \(\left(p - \frac{7}{2}\right)^2 = \frac{9}{4}\) | A1f | 1.1 |
| \(p = 2\) or 5 | A1f | 1.1 |
| [3] |
Notes
M1: Attempt \(AB^2 = BP^2 + PA^2\) (involving \(p\)) ft their \(AB\)
A1f: Correct simplified equation, ft their (a), ie:
or \(p^2 - 7p + \dfrac{46 - \text{their } a^2}{2} = 0\) or \(\left(p - \frac{7}{2}\right)^2 = \dfrac{\text{their } a^2 - 17}{4}\)
A1f: ft only their (a)
Alternative methods for M1
| Scheme | Marks |
|---|---|
| Attempt \(|PC|^2 = (\text{their radius})^2\) or \(\left(\frac{7}{2} - p\right)^2 + 4 + \frac{1}{4} = \frac{13}{2}\) | M1 |
| Attempt \(\overline{PA}\,.\,\overline{PB} = 0\) or \(((3 - p)\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \bullet ((4 - p)\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = 0\) | M1 |