M1 June 2007 Q7
7. A boat \(B\) is moving with constant velocity. At noon, \(B\) is at the point with position vector \((3\mathbf{i} - 4\mathbf{j})\) km with respect to a fixed origin \(O\). At 1430 on the same day, \(B\) is at the point with position vector \((8\mathbf{i} + 11\mathbf{j})\) km.
(a) Find the velocity of \(B\), giving your answer in the form \(p\mathbf{i} + q\mathbf{j}\). (3)
At time \(t\) hours after noon, the position vector of \(B\) is \(\mathbf{b}\) km.
(b) Find, in terms of \(t\), an expression for \(\mathbf{b}\). (3)
Another boat \(C\) is also moving with constant velocity. The position vector of \(C\), \(\mathbf{c}\) km, at time \(t\) hours after noon, is given by
\[\mathbf{c} = (-9\mathbf{i} + 20\mathbf{j}) + t(6\mathbf{i} + \lambda\mathbf{j}),\]where \(\lambda\) is a constant. Given that \(C\) intercepts \(B\),
(c) find the value of \(\lambda\), (5)
(d) show that, before \(C\) intercepts \(B\), the boats are moving with the same speed. (3)
| Scheme | Marks |
|---|---|
| \(\mathbf{v} = \dfrac{8\mathbf{i} + 11\mathbf{j} - (3\mathbf{i} - 4\mathbf{j})}{2.5}\) or any equivalent | M1 A1 |
| \(\mathbf{v} = 2\mathbf{i} + 6\mathbf{j}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathbf{b} = 3\mathbf{i} - 4\mathbf{j} + \mathbf{v}t\) ft their \(\mathbf{v}\) | M1 A1 ft |
| \(= 3\mathbf{i} - 4\mathbf{j} + (2\mathbf{i} + 6\mathbf{j})t\) | A1cao |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathbf{i}\) component: \(-9 + 6t = 3 + 2t\) | M1 |
| \(t = 3\) | M1 A1 |
| \(\mathbf{j}\) component: \(20 + 3\lambda = -4 + 18\) | M1 |
| \(\lambda = -2\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(v_B = \sqrt{(2^2 + 6^2)}\) or \(v_C = \sqrt{\left(6^2 + (-2)^2\right)}\) | M1 |
| Both correct | A1 |
| The speeds of \(B\) and \(C\) are the same cso | A1 |
| (3) | |
| (14 marks) |