AS October 2021 Paper 1 Q9
9 The points \(P(3, 5, -21)\) and \(Q(-1, 3, -16)\) are on the ceiling of a long straight underground tunnel. A ventilation shaft must be dug from the point \(M\) on the ceiling of the tunnel midway between \(P\) and \(Q\) to horizontal ground level (where the \(z\)-coordinate is 0). The ventilation shaft must be perpendicular to the tunnel.
The path of the ventilation shaft is modelled by the vector equation \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), where \(\mathbf{a}\) is the position vector of \(M\).
You are given that \(\mathbf{b} = \begin{pmatrix} 1 \\ s \\ t \end{pmatrix}\) where \(s\) and \(t\) are real numbers.
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{PQ} = \begin{pmatrix} -1 \\ 3 \\ -16 \end{pmatrix} - \begin{pmatrix} 3 \\ 5 \\ -21 \end{pmatrix} = \begin{pmatrix} -4 \\ -2 \\ 5 \end{pmatrix}\) | M1 | 2.1 |
| \(\begin{pmatrix} -4 \\ -2 \\ 5 \end{pmatrix}.\begin{pmatrix} 1 \\ s \\ t \end{pmatrix} = 0\) | M1 | 1.1 |
| \(-4 - 2s + 5t = 0\) \(\Rightarrow 2s = 5t - 4\) \(\Rightarrow s = 2.5t - 2\) | A1 | 2.1 |
| [3] |
Notes
M1: (1st) Attempt to find the direction vector of the tunnel.
Any non-zero multiple.
M1: (2nd) Use of \(\overrightarrow{PQ}.\mathbf{b} = 0\) in the solution.
A1: AG. Some intermediate work must be seen.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{M} = \tfrac{1}{2}\left(\begin{pmatrix} -1 \\ 3 \\ -16 \end{pmatrix} + \begin{pmatrix} 3 \\ 5 \\ -21 \end{pmatrix}\right) = \begin{pmatrix} 1 \\ 4 \\ -18.5 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ -18.5 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ s \\ t \end{pmatrix}\) when \(z = 0\) \(\Rightarrow -18.5 + \lambda t = 0\) | M1 | 3.4 |
| \(\Rightarrow \lambda = \dfrac{18.5}{t}\) (so \(c = 18.5\)) | A1 | 1.1 |
| [3] |
Notes
B1: Position vector (or co-ordinates) of mid-point found
M1: Using \(z = 0\) and the equation of the line to find a ‘horizontal’ relationship between \(\lambda\) and \(t\).
Condone errors in, or omission of, \(x\) and \(y\) components.
A1: NB: Question can be answered just by considering the \(z\) coordinate. If done correctly and M1 A1 gained also allow B1 as implied.
| Scheme | Marks | AO |
|---|---|---|
| So we need to minimise \(\left|\dfrac{18.5}{t}\begin{pmatrix} 1 \\ 2.5t - 2 \\ t \end{pmatrix}\right|\) | M1 | 3.3 |
| \((y =)\ \dfrac{1369}{4t^2}\left(1 + (2.5t - 2)^2 + t^2\right)\) \(= \dfrac{1369}{4}\left(7.25 - 10t^{-1} + 5t^{-2}\right)\) | M1* | 1.1 |
| So to minimise set \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{1369}{4}\left(10t^{-2} - 10t^{-3}\right) = 0\) | dep M1* | 3.1a |
| \(10t^{-2} - 10t^{-3} = 0 \Rightarrow t = 1\) | A1 | 2.2a |
| So length of shaft \(= \left|18.5\begin{pmatrix} 1 \\ 0.5 \\ 1 \end{pmatrix}\right|\) or \(\sqrt{\dfrac{1369}{4}\left(7.25 - 10 \times 1^{-1} + 5 \times 1^{-2}\right)}\) oe | M1 | 3.4 |
| \(= 18.5 \times 1.5 = 27.75\) | A1 | 1.1 |
| [6] |
Notes
M1: (1st) Stating or implying that the length of the shaft is given by \(|\lambda\mathbf{b}|\) and using their \(\lambda\) / \(t\) relationship to reduce length of shaft to a form with only one variable.
Or eg \(\left|\dfrac{18.5}{0.4s + 0.8}\begin{pmatrix} 1 \\ s \\ 0.4s + 0.8 \end{pmatrix}\right|\)
M1*: Finding expression for (squared) length of their vector
May see \(\dfrac{37}{2}\left(7.25 - 10t^{-1} + 5t^{-2}\right)^{\frac{1}{2}}\)
Or \(\dfrac{39701}{16} - \dfrac{6845}{2}t^{-1} + \dfrac{6845}{4}t^{-2}\) oe
dep M1*: Correct method for minimisation of (squared) length of their vector (eg differentiating and setting to 0)
Or attempt to complete the square in \(t^{-1}\).
\(y = \dfrac{1369}{4}\left(5(t^{-1} - 1)^2 + 2.25\right)\)
A1: (1st) So min when \(t^{-1} - 1 = 0\), \(t = 1\)
M1: (2nd) Substituting their \(t\) into their form for length of shaft
Alternate method
| Scheme | Marks |
|---|---|
| \(\mathbf{a} = \dfrac{1}{2}\left(\begin{pmatrix} -1 \\ 3 \\ -16 \end{pmatrix} + \begin{pmatrix} 3 \\ 5 \\ -21 \end{pmatrix}\right) = \begin{pmatrix} 1 \\ 4 \\ -18.5 \end{pmatrix}\) | B1 |
| \(\mathbf{n} = \begin{pmatrix} -4 \\ -2 \\ 5 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 0 \end{pmatrix} = 2\begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}\) | M1 |
| \((k)\mathbf{b} = \begin{pmatrix} -2 \\ 4 \\ 0 \end{pmatrix} \times \begin{pmatrix} -4 \\ -2 \\ 5 \end{pmatrix} = \begin{pmatrix} 20 \\ 10 \\ 20 \end{pmatrix} = 20\begin{pmatrix} 1 \\ \dfrac{1}{2} \\ 1 \end{pmatrix}\) | M1 |
| Need \(-18.5 + \lambda = 0 \Rightarrow \lambda = 18.5\) | M1 |
| So length of shaft \(= \left|18.5\begin{pmatrix} 1 \\ \dfrac{1}{2} \\ 1 \end{pmatrix}\right|\) | M1 |
| \(= 18.5 \times 3/2 = 27.75\) | A1 |
M1: (1st) Attempt to find normal to vertical plane containing tunnel
M1: (2nd) Attempt to find (multiple of) \(\mathbf{b}\) by crossing their \(\mathbf{n}\) with direction vector of tunnel.
M1: (3rd) Using \(z = 0\) to find \(\lambda\)
May see multiple of \(\mathbf{b}\) used eg \(-18.5 + 2\lambda = 0\)
M1: (4th) May see eg \(\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ -18.5 \end{pmatrix} + 9.25\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 19.5 \\ 13.25 \\ 0 \end{pmatrix}\) and then \(\begin{pmatrix} 19.5 \\ 13.25 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ -18.5 \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| So \(\mathbf{b}\) is not parallel to the \(z\)-axis so the ventilation shaft does not go straight down. | B1 | 3.2a |
| [1] |
Notes
B1: Shaft not vertical