# Question 9:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{PQ} = \begin{pmatrix}-1\\3\\-16\end{pmatrix} - \begin{pmatrix}3\\5\\-21\end{pmatrix} = \begin{pmatrix}-4\\-2\\5\end{pmatrix}$ | M1 | 2.1 — Attempt to find the direction vector of the tunnel. Any non-zero multiple. |
| $\begin{pmatrix}-4\\-2\\5\end{pmatrix}\cdot\begin{pmatrix}1\\s\\t\end{pmatrix}=0$ | M1 | 1.1 — Use of $\overrightarrow{PQ}\cdot\mathbf{b}=0$ in the solution |
| $-4-2s+5t=0 \Rightarrow 2s=5t-4 \Rightarrow s=2.5t-2$ | A1 | 2.1 — AG. Some intermediate work must be seen. |

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M = \frac{1}{2}\left(\begin{pmatrix}-1\\3\\-16\end{pmatrix}+\begin{pmatrix}3\\5\\-21\end{pmatrix}\right) = \begin{pmatrix}1\\4\\-18.5\end{pmatrix}$ | B1 | 1.1 — Position vector (or coordinates) of mid-point found |
| $\mathbf{r} = \begin{pmatrix}1\\4\\-18.5\end{pmatrix}+\lambda\begin{pmatrix}1\\s\\t\end{pmatrix}$ when $z=0$, $\Rightarrow -18.5+\lambda t=0$ | M1 | 3.4 — 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. |
| $\Rightarrow \lambda = \dfrac{18.5}{t}$ (so $c=18.5$) | A1 | 1.1 — NB: Question can be answered just by considering the $z$ coordinate. If done correctly and M1 A1 gained also allow B1 as implied. |

## Question (c):

| Answer | Mark | Guidance |
|--------|------|----------|
| So we need to minimise $\left\|\frac{18.5}{t}\begin{pmatrix}1\\2.5t-2\\t\end{pmatrix}\right\|$ | **M1** | 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 e.g. $\left\|\frac{18.5}{0.4s+0.8}\begin{pmatrix}1\\s\\0.4s+0.8\end{pmatrix}\right\|$ |
| $(y=)\frac{1369}{4t^2}(1+(2.5t-2)^2+t^2) = \frac{1369}{4}(7.25-10t^{-1}+5t^{-2})$ | **M1\*** | Finding expression for (squared) length of their vector. May see $\frac{37}{2}(7.25-10t^{-1}+5t^{-2})^{\frac{1}{2}}$ or $\frac{39701}{16}-\frac{6845}{2}t^{-1}+\frac{6845}{4}t^{-2}$ |
| So to minimise set $\frac{dy}{dt}=\frac{1369}{4}(10t^{-2}-10t^{-3})=0$ | **dep M1\*** | Correct method for minimisation of (squared) length of their vector (e.g. differentiating and setting to 0). Or attempt to complete the square in $t^{-1}$: $y=\frac{1369}{4}(5(t^{-1}-1)^2+2.25)$ |
| $10t^{-2}-10t^{-3}=0 \Rightarrow t=1$ | **A1** | So min when $t^{-1}-1=0$, $t=1$ |
| So length of shaft $= \left\|18.5\begin{pmatrix}1\\0.5\\1\end{pmatrix}\right\|$ or $\sqrt{\frac{1369}{4}(7.25-10\times1^{-1}+5\times1^{-2})}$ | **M1** | Substituting their $t$ into their form for length of shaft |
| $= 18.5 \times 1.5 = 27.75$ | **A1** | |

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## Alternate Method for (c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{a}=\frac{1}{2}\left(\begin{pmatrix}-1\\3\\-16\end{pmatrix}+\begin{pmatrix}3\\5\\-21\end{pmatrix}\right)=\begin{pmatrix}1\\4\\-18.5\end{pmatrix}$ | **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** | Attempt to find normal to vertical plane containing tunnel |
| $(k)\mathbf{b}=\begin{pmatrix}-2\\4\\0\end{pmatrix}\times\begin{pmatrix}-4\\-2\\5\end{pmatrix}=\begin{pmatrix}20\\10\\20\end{pmatrix}=20\begin{pmatrix}1\\\frac{1}{2}\\1\end{pmatrix}$ | **M1** | Attempt to find (multiple of) $\mathbf{b}$ by crossing their $\mathbf{n}$ with direction vector of tunnel |
| Need $-18.5+\lambda=0 \Rightarrow \lambda=18.5$ | **M1** | Using $z=0$ to find $\lambda$. May see multiple of $\mathbf{b}$ used e.g. $-18.5+2\lambda=0$ |
| So length of shaft $= \left\|18.5\begin{pmatrix}1\\\frac{1}{2}\\1\end{pmatrix}\right\|$ | **M1** | May see e.g. $\mathbf{r}=\begin{pmatrix}1\\4\\-18.5\end{pmatrix}+9.25\begin{pmatrix}1\\1\\2\end{pmatrix}=\begin{pmatrix}19.5\\13.25\\0\end{pmatrix}$ and then $\begin{pmatrix}19.5\\13.25\\0\end{pmatrix}-\begin{pmatrix}1\\4\\-18.5\end{pmatrix}$ |
| $= 18.5 \times \frac{3}{2} = 27.75$ | **A1** | |

**[6]**

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## Question (d):

| Answer | Mark | Guidance |
|--------|------|----------|
| So $\mathbf{b}$ is not parallel to the $z$-axis so the ventilation shaft does not go straight down. | **B1** | Shaft not vertical |

**[1]**