AS June 2018 Paper 1 Q4
4. Part of the mains water system for a housing estate consists of water pipes buried beneath the ground surface. The water pipes are modelled as straight line segments. One water pipe, \(W\), is buried beneath a particular road. With respect to a fixed origin \(O\), the road surface is modelled as a plane with equation \(3x - 5y - 18z = 7\), and \(W\) passes through the points \(A(-1, -1, -3)\) and \(B(1, 2, -3)\). The units are in metres.
A point \(C(-1, -2, 0)\) lies on the road. A section of water pipe needs to be connected to \(W\) from \(C\).
| Scheme | Marks | AO |
|---|---|---|
| Attempts the scalar product between the direction of \(W\) and the normal to the road and uses trigonometry to find an angle. | M1 | 3.1a |
| \[\left(\begin{pmatrix}1\\ 2\\ -3\end{pmatrix} - \begin{pmatrix}-1\\ -1\\ -3\end{pmatrix}\right)\bullet\begin{pmatrix}3\\ -5\\ -18\end{pmatrix} = -9\ \text{ or }\ \left(\begin{pmatrix}-1\\ -1\\ -3\end{pmatrix} - \begin{pmatrix}1\\ 2\\ -3\end{pmatrix}\right)\bullet\begin{pmatrix}3\\ -5\\ -18\end{pmatrix} = 9\] | M1 A1 | 1.1b 1.1b |
| \[\sqrt{(2)^2 + (3)^2 + (0)^2}\sqrt{(3)^2 + (-5)^2 + (-18)^2}\cos\alpha = \text{“}{-9}\text{”}\]\[\theta = 90 - \arccos\left(\frac{9}{\sqrt{13}\sqrt{358}}\right) \text{ or } \theta = \arcsin\left(\frac{9}{\sqrt{13}\sqrt{358}}\right)\]Angle between pipe and road \(= 7.58^\circ\) (3sf) or 0.132 radians (3sf) (Allow \(-7.58^\circ\) or \(-0.132\) radians) | M1 A1 | 1.1b 3.2a |
| (5) |
Notes
(Corrected from the printed mark scheme: the printed working has \((3)^3\) and \((-5)^3\) inside the square roots; these are squares, \((3)^2\) and \((-5)^2\), giving \(\sqrt{13}\) and \(\sqrt{358}\).)
M1: Realises the scalar product between the direction of \(W\) and the normal to the road is needed and so applies it and uses trigonometry to find an angle
M1: Calculates the scalar product between \(\pm\left(\begin{pmatrix}1\\ 2\\ -3\end{pmatrix} - \begin{pmatrix}-1\\ -1\\ -3\end{pmatrix}\right)\) and \(\pm\begin{pmatrix}3\\ -5\\ -18\end{pmatrix}\) (Allow sign slips as long as the intention is clear)
A1: \(\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\bullet\begin{pmatrix}3\\ -5\\ -18\end{pmatrix} = -9\) or \(\begin{pmatrix}-2\\ -3\\ 0\end{pmatrix}\bullet\begin{pmatrix}3\\ -5\\ -18\end{pmatrix} = 9\) or \(\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\bullet\begin{pmatrix}-3\\ 5\\ 18\end{pmatrix} = 9\) or \(\begin{pmatrix}-2\\ -3\\ 0\end{pmatrix}\bullet\begin{pmatrix}-3\\ 5\\ 18\end{pmatrix} = -9\)
M1: A fully complete and correct method for obtaining the acute angle
A1: Awrt \(7.58^\circ\) or awrt 0.132 radians (must see units). Do not isw and withhold this mark if extra answers are given.
| Scheme | Marks | AO |
|---|---|---|
| \[W: \begin{pmatrix}-1\\ -1\\ -3\end{pmatrix} + t\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\ \text{ or }\ \begin{pmatrix}1\\ 2\\ -3\end{pmatrix} + \lambda\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\] | B1ft | 1.1b |
| \[C \text{ to } W: \left\{\begin{pmatrix}-1\\ -1\\ -3\end{pmatrix} + t\begin{pmatrix}2\\ 3\\ 0\end{pmatrix} - \begin{pmatrix}-1\\ -2\\ 0\end{pmatrix}\right\}\ \text{ or }\ \left\{\begin{pmatrix}1\\ 2\\ -3\end{pmatrix} + \lambda\begin{pmatrix}2\\ 3\\ 0\end{pmatrix} - \begin{pmatrix}-1\\ -2\\ 0\end{pmatrix}\right\}\] | M1 | 3.4 |
| \[\begin{pmatrix}2t\\ 3t + 1\\ -3\end{pmatrix}\bullet\begin{pmatrix}2\\ 3\\ 0\end{pmatrix} = 0 \Rightarrow t = \ldots\ \text{ or }\ \begin{pmatrix}2 + 2\lambda\\ 4 + 3\lambda\\ -3\end{pmatrix}\bullet\begin{pmatrix}2\\ 3\\ 0\end{pmatrix} = 0 \Rightarrow \lambda = \ldots\]or\[(2t)^2 + (3t + 1)^2 + (-3)^2 = \ldots \text{ or } (2 + 2t)^2 + (4 + 3t)^2 + (-3)^2 = \ldots\] | M1 | 3.1b |
| \[t = -\frac{3}{13} \text{ or } \lambda = -\frac{16}{13} \Rightarrow (C \text{ to } W)_{\min} \text{ is } -\frac{6}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} - 3\mathbf{k}\]or\[(2t)^2 + (3t + 1)^2 + (-3)^2 = 13\left(t + \frac{3}{13}\right)^2 + \frac{121}{13}\]or\[(2 + 2t)^2 + (4 + 3t)^2 + (-3)^2 = 13\left(\lambda + \frac{16}{13}\right)^2 + \frac{121}{13}\]or\[\frac{\mathrm{d}\left((2t)^2 + (3t + 1)^2 + (-3)^2\right)}{\mathrm{d}t} = 0 \Rightarrow t = -\frac{3}{13} \Rightarrow C \text{ to } W \text{ is } -\frac{6}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} - 3\mathbf{k}\]Or\[\frac{\mathrm{d}\left((2 + 2t)^2 + (4 + 3t)^2 + (-3)^2\right)}{\mathrm{d}t} = 0 \Rightarrow t = -\frac{16}{13} \Rightarrow (C \text{ to } W)_{\min} \text{ is } -\frac{6}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} - 3\mathbf{k}\] | A1 | 1.1b |
| \[d = \sqrt{\left(-\frac{6}{13}\right)^2 + \left(\frac{4}{13}\right)^2 + (-3)^2} \text{ or } d = \sqrt{\frac{121}{13}}\] | ddM1 | 1.1b |
| Shortest length of pipe needed is 305 or 305 cm or 3.05 m | A1 | 3.2a |
| (6) | ||
| (11 marks) |
Notes
B1ft: Forms the correct parametric form for the pipe \(W\). Follow through their direction vector for \(W\) from part (a).
M1: Identifies the need to and forms a vector connecting \(C\) to \(W\) using a parametric form for \(W\)
M1: Uses the model to form the scalar product of \(C\) to \(W\) and the direction of \(W\) to find the value of their parameter or finds the distance \(C\) to \(W\) or \((C \text{ to } W)^2\) in terms of their parameter
A1: Correct vector or correct completion of the square
ddM1: Correct use of Pythagoras on their vector \(CW\) or appropriate method to find the shortest distance between the point and the pipe. Dependent on both previous method marks.
A1: Correct length for the required section of pipe is 305 or 305 cm or 3.05 m
Alternatives for part (b): Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{AC} = \begin{pmatrix}0\\ 1\\ -3\end{pmatrix},\quad \mathbf{AB} = \begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\) | B1ft | 1.1b |
| \(\mathbf{AC}.\mathbf{AB} = \begin{pmatrix}0\\ 1\\ -3\end{pmatrix}.\begin{pmatrix}2\\ 3\\ 0\end{pmatrix} = 3\) | M1 | 3.4 |
| \(\Rightarrow \cos CAB = \dfrac{3}{\sqrt{10}\sqrt{13}} \Rightarrow CAB = \ldots\) | M1 | 3.1b |
| \(CAB = 74.74\ldots^\circ\) | A1 | 1.1b |
| \(d = \sqrt{10}\sin 74.74\ldots^\circ\) | ddM1 | 1.1b |
| Shortest length of pipe needed is 305 or 305 cm or 3.05 m | A1 | 3.2a |
| (6) |
(Note: the vector \(\begin{pmatrix}0\\ 1\\ -3\end{pmatrix}\) printed here is \(\overrightarrow{CA}\); \(\overrightarrow{AC} = \begin{pmatrix}0\\ -1\\ 3\end{pmatrix}\), which gives a scalar product of \(-3\) and angle \(CAB = 105.25\ldots^\circ\). Since \(\sin 105.25\ldots^\circ = \sin 74.74\ldots^\circ\), the distance is the same.)
B1ft: Forms the correct vectors. Follow through their direction vector for \(W\) from part (a).
M1: Identifies the need to and forms the scalar product between AC and AB
M1: Uses the model to form the scalar product and uses this to find the angle \(CAB\)
A1: Correct angle
ddM1: Correct method using their values or appropriate method to find the shortest distance between the point and the pipe. Dependent on both previous method marks.
A1: Correct length for the required section of pipe is 305 or 305 cm or 3.05 m
Way 3
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{AC} = \begin{pmatrix}0\\ -1\\ 3\end{pmatrix},\quad \mathbf{AB} = \begin{pmatrix}2\\ 3\\ 0\end{pmatrix}\) | B1ft | 1.1b |
| \(\mathbf{AC} \times \mathbf{AB} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & -1 & 3\\ 2 & 3 & 0\end{vmatrix} = \begin{pmatrix}-9\\ 6\\ 2\end{pmatrix}\) | M1 | 3.4 |
| \(|\mathbf{AC} \times \mathbf{AB}| = \sqrt{9^2 + 6^2 + 2^2} = \ldots\) | M1 | 3.1b |
| \(= 11\) | A1 | 1.1b |
| \(d = \dfrac{11}{|\mathbf{AB}|} = \dfrac{11}{\sqrt{2^2 + 3^2}} = \ldots\) | ddM1 | 1.1b |
| Shortest length of pipe needed is 305 or 305 cm or 3.05 m | A1 | 3.2a |
| (6) |
(Corrected from the printed mark scheme: Way 3 prints \(\mathbf{AC} = \begin{pmatrix}0\\ 1\\ -3\end{pmatrix}\); \(\mathbf{AC} = \begin{pmatrix}0\\ -1\\ 3\end{pmatrix}\), as used in the vector product.)
B1ft: Forms the correct vectors. Follow through their direction vector for \(W\) from part (a).
M1: Identifies the need to and forms the vector product between AC and AB
M1: Uses the model to find the magnitude of their vector product
A1: Correct value
ddM1: Correct method using their values or appropriate method to find the shortest distance between the point and the pipe. Dependent on both previous method marks.
A1: Correct length for the required section of pipe is 305 or 305 cm or 3.05 m