# Question 8:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{b} = \pm\left[\begin{pmatrix}300\\300\\-50\end{pmatrix} - \begin{pmatrix}-300\\400\\-150\end{pmatrix}\right] = \pm\begin{pmatrix}600\\-100\\100\end{pmatrix}$ | M1 | Attempts direction between positions $P$ and $Q$; two correct entries imply method |
| $\mathbf{r} = \begin{pmatrix}-300\\400\\-150\end{pmatrix} + \lambda\begin{pmatrix}600\\-100\\100\end{pmatrix}$ | A1 | Correct equation in correct form; any point on line may be used, any non-zero multiple of direction; must begin $\mathbf{r} = \ldots$ |

## Part (b)(i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $k = 200$ | B1 | Correct value of $k$ deduced |
| $\overrightarrow{MX} = \begin{pmatrix}-300\\400\\-150\end{pmatrix} + \lambda\begin{pmatrix}600\\-100\\100\end{pmatrix} - \begin{pmatrix}100\\k\\100\end{pmatrix} = \begin{pmatrix}-400+600\lambda\\200-100\lambda\\-250+100\lambda\end{pmatrix}$ | M1 | Realises need to find distance from point on mountain to general point on line |
| $\begin{pmatrix}-400+600\lambda\\200-100\lambda\\-250+100\lambda\end{pmatrix} \cdot \begin{pmatrix}600\\-100\\100\end{pmatrix} = 0 \Rightarrow \lambda = \ldots$ | dM1 | Takes dot product with direction vector, sets to zero and solves for $\lambda$; allow finding either $\lambda$ in terms of $k$ or $k$ in terms of $\lambda$ |
| $\overrightarrow{OX} = \begin{pmatrix}-300\\400\\-150\end{pmatrix} + \frac{3}{4}\begin{pmatrix}600\\-100\\100\end{pmatrix} = \ldots$ | M1 | Substitutes $\lambda$ into line equation |
| Coordinates of $X$ are $(150, 325, -75)$ | A1 | Correct point |

## Part (b)(ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Length of tunnel $= \sqrt{(150-100)^2 + (325-200)^2 + (-75-100)^2} = \ldots$ | M1 | Uses distance formula with their point and $M$, or with $\overrightarrow{MX}$ from (i) |
| $\approx 221$ m | A1 | Correct distance including units; accept $221$ m or $25\sqrt{78}$ m |

## Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $|\overrightarrow{OP}| = \sqrt{(-300)^2 + 400^2 + (-150)^2} \approx 522$ and $|\overrightarrow{OQ}| = \sqrt{300^2 + 300^2 + 50^2} \approx 427$ | M1 | Calculates the two distances $OP$ and $OQ$ |
| New tunnel length significantly shorter so company will likely build the accessway | A1ft | Appropriate conclusion for tunnel length; distances $OP$ and $OQ$ must be correct; reason and conclusion needed; accept "significantly shorter" or "tunnel more than 100m less than either existing accessway" |

## Part (d):

| Working/Answer | Mark | Guidance |
|---|---|---|
| E.g. mountainside is not likely to be flat so a plane may not be a good model; tunnel/pipeline will not have negligible thickness so modelling as lines may not be appropriate; a shortest length tunnel may not be possible as strata of rock have not been considered | B1 | Any appropriate criticism referring to the model; must reference model in some way; note: reference to measurements not being correct is **NOT** a limitation |

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