M1 June 2011 Q7
7. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north respectively. Position vectors are given relative to a fixed origin \(O\).]
Two ships \(P\) and \(Q\) are moving with constant velocities. Ship \(P\) moves with velocity \((2\mathbf{i} - 3\mathbf{j})\) km h\(^{-1}\) and ship \(Q\) moves with velocity \((3\mathbf{i} + 4\mathbf{j})\) km h\(^{-1}\).
(a) Find, to the nearest degree, the bearing on which \(Q\) is moving. (2)
At 2 pm, ship \(P\) is at the point with position vector \((\mathbf{i} + \mathbf{j})\) km and ship \(Q\) is at the point with position vector \((-2\mathbf{j})\) km.
At time \(t\) hours after 2 pm, the position vector of \(P\) is \(\mathbf{p}\) km and the position vector of \(Q\) is \(\mathbf{q}\) km.
(b) Write down expressions, in terms of \(t\), for
(i) \(\mathbf{p}\),
(ii) \(\mathbf{q}\),
(iii) \(\overrightarrow{PQ}\). (5)
(c) Find the time when
(i) \(Q\) is due north of \(P\),
(ii) \(Q\) is north-west of \(P\). (4)
| Scheme | Marks |
|---|---|
| \(\tan\theta = \tfrac{3}{4}\); bearing is 37\(^\circ\) (nearest degree) | M1; A1 |
| (2) |
| Scheme | Marks |
|---|---|
| (i) \(\mathbf{p} = (\mathbf{i} + \mathbf{j}) + t(2\mathbf{i} - 3\mathbf{j})\) | M1 A1 |
| (ii) \(\mathbf{q} = (-2\mathbf{j}) + t(3\mathbf{i} + 4\mathbf{j})\) | A1 |
| (iii) \(\mathbf{PQ} = \mathbf{q} - \mathbf{p} = (-\mathbf{i} - 3\mathbf{j}) + t(\mathbf{i} + 7\mathbf{j})\) | M1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| (i) \(-1 + t = 0\) | M1 |
| \(t = 1\) or 3pm | A1 |
| (ii) \(-1 + t = -(-3 + 7t)\) | M1 |
| \(t = \tfrac{1}{2}\) or 2.30 pm | A1 |
| (4) | |
| (11 marks) |