M1 June 2017 Q7
7. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin \(O\).]
Two ships, \(P\) and \(Q\), are moving with constant velocities.
The velocity of \(P\) is \((9\mathbf{i} - 2\mathbf{j})\) km h\(^{-1}\) and the velocity of \(Q\) is \((4\mathbf{i} + 8\mathbf{j})\) km h\(^{-1}\)
When \(t = 0\), the position vector of \(P\) is \((9\mathbf{i} + 10\mathbf{j})\) km and the position vector of \(Q\) is \((\mathbf{i} + 4\mathbf{j})\) km. At time \(t\) hours, the position vectors of \(P\) and \(Q\) are \(\mathbf{p}\) km and \(\mathbf{q}\) km respectively.
| Scheme | Marks |
|---|---|
| \(\tan\theta = \tfrac{2}{9}\ \ \ \theta = 12.5^\circ\ \ \ \) bearing \(103^\circ\) | M1 A1 A1 |
| (3) |
Notes
M1 for \(\tan\theta = \pm\tfrac{2}{9}\) or \(\pm\tfrac{9}{2}\) or use \(\sin\theta\) or \(\cos\theta\)
First A1 for \(\theta = \pm 13^\circ\) or \(\pm 77^\circ\) or \(\pm 12.5^\circ\) or \(\pm 77.5^\circ\) or better
Second A1 for \(103^\circ\)
| Scheme | Marks |
|---|---|
| \(\mathbf{p} = (9\mathbf{i} + 10\mathbf{j}) + t(9\mathbf{i} - 2\mathbf{j})\) | M1 A1 |
| \(\mathbf{q} = (\mathbf{i} + 4\mathbf{j}) + t(4\mathbf{i} + 8\mathbf{j})\) | A1 |
| (3) |
Notes
M1 for clear attempt at \(\mathbf{p} = (9\mathbf{i} + 10\mathbf{j}) + t(9\mathbf{i} - 2\mathbf{j})\) or \(\mathbf{q} = (\mathbf{i} + 4\mathbf{j}) + t(4\mathbf{i} + 8\mathbf{j})\) (Allow slips but must be a ‘+’ sign and \(\mathbf{r} + t\,\mathbf{v}\))
(i) First A1 for \(\mathbf{p} = (9\mathbf{i} + 10\mathbf{j}) + t(9\mathbf{i} - 2\mathbf{j})\) oe
(ii) Second A1 for \(\mathbf{q} = (\mathbf{i} + 4\mathbf{j}) + t(4\mathbf{i} + 8\mathbf{j})\) oe
| Scheme | Marks |
|---|---|
| \(\overrightarrow{QP} = (8 + 5t)\mathbf{i} + (6 - 10t)\mathbf{j}\) | M1 A1 |
| (2) |
Notes
M1 for \(\mathbf{p} - \mathbf{q}\) or \(\mathbf{q} - \mathbf{p}\) with their \(\mathbf{p}\) and \(\mathbf{q}\) substituted
A1 for correct answer \(\overrightarrow{QP} = (8 + 5t)\mathbf{i} + (6 - 10t)\mathbf{j}\) (don’t need \(\overrightarrow{QP}\) but on R.H.S must be identical coefficients of \(\mathbf{i}\) and \(\mathbf{j}\) but allow column vectors)
| Scheme | Marks |
|---|---|
| \(D^2 = (8 + 5t)^2 + (6 - 10t)^2\) | M1 |
| \(= 125t^2 - 40t + 100\) | A1 |
| \(100 = 125t^2 - 40t + 100\) | M1 |
| \(0 = 5t(25t - 8)\) | M1 |
| \(t = 0\) or 0.32 | A1 A1 |
| (6) | |
| (14 marks) |
Notes
First M1 for attempt to find \(QP\) or \(QP^2\) in terms of \(t\) only, using correct formula
First A1 for a correct expression (with or without \(\sqrt{\ }\)) \(125t^2 - 40t + 100\)
Second M1 for \(\sqrt{\ }\)(3 term quadratic) = 10 or (3 term quadratic) = 100.
Third M1 for quadratic expression = 0 and attempt to solve (e.g. factorising or using formula)
Second A1 for \(t = 0\) (if they divide by \(t\) and lose this value but get 0.32, M1A0A1)
Third A1 for \(t = 0.32\) oe