AS June 2025 Paper 1 Q9
9. An engineer detects a source of water below the surface of the ground. The engineer models the situation relative to a fixed origin \(O\).
In the model
- the surface of the ground is a plane \(\Pi\) with equation \(x - 2y + 8z = 1\)
- the source of water is at a point \(W\) with coordinates \((6, -2, -4)\)
where the units are metres.
To access the water, a hole is drilled, in a straight line, from a point \(P\) on the surface of the ground to \(W\).
Given that the length of the hole needs to be as short as possible,
Given that the actual length of the hole drilled is 2.52 metres,
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{(1 \times 6) + (-2 \times -2) + (8 \times -4) \pm 1}{\sqrt{1^2 + (-2)^2 + 8^2}}\) | M1 | 3.4 |
| \(\dfrac{23}{\sqrt{69}}\) or \(\dfrac{1}{3}\sqrt{69}\) or awrt 2.77 | A1 | 1.1b |
| (2) |
Notes
M1: Uses the shortest distance formula \(\dfrac{(1 \times 6) + (-2 \times -2) + (8 \times -4) \pm 1}{\sqrt{1^2 + (-2)^2 + 8^2}}\) condoning a sign slip with \(d\) and no modulus for this mark.
May find the answer to (c) first and then find the distance. It must be a correct method, scoring at least M1M1 in (c)
A1: Correct exact answer or awrt 2.77 isw
| Scheme | Marks | AO |
|---|---|---|
| e.g. not reliable as The source of the water will not be at a single point The water level will vary The ground may not be flat | B1 | 3.2b |
| (1) |
Notes
B1: See scheme
Note: the position of the water may change is B0
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{r} = \begin{pmatrix}6\\ -2\\ -4\end{pmatrix} + \lambda\begin{pmatrix}1\\ -2\\ 8\end{pmatrix}\) and \((6 + \lambda) - 2(-2 - 2\lambda) + 8(-4 + 8\lambda) = 1 \Rightarrow \lambda = \ldots\left\{\dfrac{1}{3}\right\}\) Or Their \(\dfrac{1}{3}\sqrt{69}\) divided by \(\sqrt{1^2 + (-2)^2 + (8)^2}\) leading to a value for \(\lambda\) | M1 | 3.1b |
| \(\mathbf{r} = \begin{pmatrix}6\\ -2\\ -4\end{pmatrix} + \text{their}\dfrac{1}{3}\begin{pmatrix}1\\ -2\\ 8\end{pmatrix} = \ldots\) | dM1 | 1.1b |
| \(\left(\dfrac{19}{3}, -\dfrac{8}{3}, -\dfrac{4}{3}\right)\) or \(\left(6\dfrac{1}{3}, -2\dfrac{2}{3}, -1\dfrac{1}{3}\right)\) only | A1 | 1.1b |
| (3) |
Notes
(c) This may be seen in part (a) to score any marks at least the final coordinate must be stated in (c) or ‘see (a)’
M1: A complete method to find the value of \(\lambda\)
dM1: Dependent on the previous method mark. Uses their value of \(\lambda\) in the equation of the line \(\mathbf{r} = \begin{pmatrix}6\\ -2\\ -4\end{pmatrix} + \lambda\begin{pmatrix}1\\ -2\\ 8\end{pmatrix} = \ldots\) to find a coordinate
A1: Correct coordinate
| Scheme | Marks | AO |
|---|---|---|
| Compares their answer to (a) with 2.52, must give some idea about the difference between the values and draws an appropriate conclusion Answer from (a)
| B1 | 3.5a |
| (7 marks) |
Notes
B1: See scheme, must have an answer to (a) so score this mark