## Question 9:

### Part 9(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix} 9 \\ 4 \\ 11 \end{pmatrix} - \begin{pmatrix} -3 \\ 1 \\ -7 \end{pmatrix} = \begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix}$ or $\mathbf{d} = \begin{pmatrix} 4 \\ 1 \\ 6 \end{pmatrix}$ | M1 | 3.1a — Attempts to find $\overrightarrow{OB} - \overrightarrow{OA}$ or $\overrightarrow{OA} - \overrightarrow{OB}$ or direction vector **d** |
| $\overrightarrow{OF} = \mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ -7 \end{pmatrix} + \lambda \begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix}$ | M1 | 1.1b — Applies $\overrightarrow{OA} + \lambda(\overrightarrow{AB}$ or $\overrightarrow{BA}$ or **d**) |
| $\overrightarrow{OF} \cdot \overrightarrow{AB} = 0 \Rightarrow \begin{pmatrix} -3+12\lambda \\ 1+3\lambda \\ -7+18\lambda \end{pmatrix} \cdot \begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix} = 0$ | dM1 | 1.1b — Depends on previous M. Sets $\overrightarrow{OF}$ (in terms of $\lambda$) $\cdot \overrightarrow{AB} = 0$ |
| $\Rightarrow -36 + 144\lambda + 3 + 9\lambda - 126 + 324\lambda = 0 \Rightarrow 477\lambda - 159 = 0$ | | |
| $\Rightarrow \lambda = \dfrac{1}{3}$ | A1 | 1.1b — Lambda correct |
| $\overrightarrow{OF} = \begin{pmatrix} -3 \\ 1 \\ -7 \end{pmatrix} + \dfrac{1}{3}\begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and minimum distance $= \sqrt{(1)^2 + (2)^2 + (-1)^2}$ | dM1 | 3.1a — Depends on previous M. Complete method for finding $|\overrightarrow{OF}|$ |
| $= \sqrt{6}$ or $2.449\ldots$ | A1 | 1.1b |
| $> 2$, so the octopus is not able to catch the fish $F$ | A1ft | 3.2a — Correct follow through conclusion in context |

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### Part 9(a) — Alternative 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix}$ or $\mathbf{d} = \begin{pmatrix} 4 \\ 1 \\ 6 \end{pmatrix}$ | M1 | 3.1a |
| $\overrightarrow{OA} = \begin{pmatrix} -3 \\ 1 \\ -7 \end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix} \Rightarrow \overrightarrow{OA} \cdot \overrightarrow{AB}$ | M1 | 1.1b — Realisation that dot product between $\overrightarrow{OA}$ and $\overrightarrow{AB}$ is required |
| $\cos\theta = \dfrac{\overrightarrow{OA} \cdot \overrightarrow{AB}}{|\overrightarrow{OA}||\overrightarrow{AB}|} = \dfrac{\pm\left(\begin{pmatrix}-3\\1\\-7\end{pmatrix} \cdot \begin{pmatrix}12\\3\\18\end{pmatrix}\right)}{\sqrt{(-3)^2+(1)^2+(-7)^2} \cdot \sqrt{(12)^2+(3)^2+(18)^2}}$ | dM1 | 1.1b — Applies dot product formula between $\overrightarrow{OA}$ and $\overrightarrow{AB}$ |
| $\cos\theta = \dfrac{-36+3-126}{\sqrt{59}\cdot\sqrt{477}} = \dfrac{-159}{\sqrt{59}\cdot\sqrt{477}}$ | | |
| $\theta = 161.4038\ldots$ or $18.5962\ldots$ or $\sin\theta = 0.3188\ldots$ | A1 | 1.1b |
| minimum distance $= \sqrt{(-3)^2+(1)^2+(-7)^2}\sin(18.5962\ldots)$ | dM1 | 3.1a — Depends on previous M. Uses $|OA|\sin\theta$ |
| $= \sqrt{6}$ or $2.449\ldots$ | A1 | 1.1b |
| $> 2$, so the octopus is not able to catch the fish $F$ | A1ft | 3.2a |

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### Part 9(a) — Alternative 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix} 12 \\ 3 \\ 18 \end{pmatrix}$ or $\mathbf{d} = \begin{pmatrix} 4 \\ 1 \\ 6 \end{pmatrix}$ | M1 | 3.1a |
| $\overrightarrow{OF} = \mathbf{r} = \begin{pmatrix}-3\\1\\-7\end{pmatrix} + \lambda\begin{pmatrix}12\\3\\18\end{pmatrix}$ | M1 | 1.1b |
| $\left|\overrightarrow{OF}\right|^2 = (-3+12\lambda)^2 + (1+3\lambda)^2 + (-7+18\lambda)^2$ | dM1 | 1.1b — Applies Pythagoras by finding $|\overrightarrow{OF}|^2$ |
| $= 9 - 72\lambda + 144\lambda^2 + 1 + 6\lambda + 9\lambda^2 + 49 - 252\lambda + 324\lambda^2$ | | |
| $= 477\lambda^2 - 318\lambda + 59$ | A1 | 1.1b |
| $= 53(3\lambda - 1)^2 + 6$ | dM1 | 3.1a — Method of completing the square or differentiating |
| minimum distance $= \sqrt{6}$ or $2.449\ldots$ | A1 | 1.1b |
| $> 2$, so the octopus is not able to catch the fish $F$ | A1ft | 3.2a |

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### Part 9(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. Fish $F$ may not swim in an exact straight line from $A$ to $B$; Fish $F$ may hit an obstacle whilst swimming from $A$ to $B$; Fish $F$ may deviate its path to avoid being caught by the octopus | B1 | 3.5b — An acceptable criticism for fish $F$, in context |

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### Part 9(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. Octopus is effectively modelled as a particle — so we may need to look at where the octopus's mass is distributed; Octopus may during the fish $F$'s motion move away from its fixed location at $O$ | B1 | 3.5b — An acceptable criticism for the octopus, in context |