# Question 12:

## Part (a):
| $(2, 0, 7)$ | B1 | Accept vector equivalent |

## Part (b):
| $\begin{pmatrix}2\\-2\\1\end{pmatrix}\cdot\begin{pmatrix}8\\4\\1\end{pmatrix} = 16-8+1 = 3\times9\cos\theta$ | M1A1 | Correct full method for scalar product of direction vectors |
| $\cos\theta = \frac{1}{3}$ | A1 | May be implied |
| $\sin^2\theta + \cos^2\theta = 1 \Rightarrow \sin\theta = \sqrt{1-\left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2}{3}\sqrt{2}$ | M1A1* | Allow right-angled triangle methods; must see correct work e.g. $\cos\theta=\frac{1}{3}\Rightarrow\sin\theta=\frac{\sqrt{3^2-1}}{3}$ |

## Part (c):
| $AB = \sqrt{8^2+8^2+4^2}$ OR $AB = 4\times3$ | M1 | |
| Area $= \frac{1}{2}ab\sin C = \frac{1}{2}\times12\times24\times\frac{2}{3}\sqrt{2} = 96\sqrt{2}$ | M1A1 | |

## Part (d):
| Attempts to find $\mu$ by $\frac{\text{length }AC}{|(8,4,1)|} = \frac{24}{9}$ | M1A1 | |
| Attempts $\begin{pmatrix}2\\0\\7\end{pmatrix} \pm \frac{8}{3}\begin{pmatrix}8\\4\\1\end{pmatrix}$ | dM1 | Dependent on previous method mark |
| $\left(\frac{70}{3},\frac{32}{3},\frac{29}{3}\right), \left(-\frac{58}{3},-\frac{32}{3},\frac{13}{3}\right)$ | A1A1 | |

# Mark Scheme Extraction

## Part (a)/(b) [Trigonometry]:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \frac{1}{3} \Rightarrow \sin\theta = \sin\left(\cos^{-1}\frac{1}{3}\right) = \frac{2}{3}\sqrt{2}$ | | Allow e.g. $\cos\theta = \frac{1}{3} \Rightarrow \theta = 70.52...\sin\theta = \sin(70.52...) = \frac{2}{3}\sqrt{2}$ or just $\cos\theta = \frac{1}{3} \Rightarrow \sin\theta = \frac{2}{3}\sqrt{2}$ |
| $\sin\theta = \sqrt{\frac{8}{9}} = \frac{2}{3}\sqrt{2}$ | A1* | With no need to state the value of $k$ |

---

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $|AB| = |\mathbf{b} - \mathbf{a}| = \sqrt{8^2 + 8^2 + 4^2}$ | M1 | A correct method of finding $|AB|$; alternatively uses $|AB| = 4\times`3`$ |
| Area $= \frac{1}{2}|AB|\times 2|AB|\sin\theta$ with their $\sin\theta$ | M1 | But not $\frac{1}{2}|AB|\times\frac{1}{2}|AB|\sin\theta$ |
| $96\sqrt{2}$ | A1 | |

---

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find $\mu$ by $\dfrac{\text{length } AC}{|(8,4,1)|}$ e.g. $\sqrt{(8\mu)^2 + (4\mu)^2 + \mu^2} = \text{``}24\text{''}$ | M1 | |
| $(8\mu)^2 + (4\mu)^2 + \mu^2 = \text{``}24\text{''}^2$ | | **but not** $\sqrt{(8\mu)^2 + (4\mu)^2 + \mu^2} = 24^2$ i.e. both sides must be consistent |
| $\mu = (\pm)\dfrac{24}{9}$ | A1 | |
| Attempts to find at least one position for $C$ by using $\begin{pmatrix}2\\0\\7\end{pmatrix} \pm \textbf{their } \frac{8}{3}\begin{pmatrix}8\\4\\1\end{pmatrix}$ | dM1 | |
| Either of $\left(\dfrac{70}{3}, \dfrac{32}{3}, \dfrac{29}{3}\right), \left(-\dfrac{58}{3}, -\dfrac{32}{3}, \dfrac{13}{3}\right)$ | A1 | Allow in vector form as $\begin{pmatrix}70/3\\32/3\\29/3\end{pmatrix}$ or $\frac{1}{3}\begin{pmatrix}70\\32\\29\end{pmatrix}$ or $\begin{pmatrix}-58/3\\-32/3\\13/3\end{pmatrix}$ or $\frac{1}{3}\begin{pmatrix}-58\\-32\\13\end{pmatrix}$ but not e.g. $\frac{1}{3}(70,32,29)$ |
| Both of $\left(\dfrac{70}{3}, \dfrac{32}{3}, \dfrac{29}{3}\right), \left(-\dfrac{58}{3}, -\dfrac{32}{3}, \dfrac{13}{3}\right)$ | A1 | Allow in vector form as above |

**Note:** Using $\overrightarrow{OC} = \overrightarrow{OA} \pm 2\overrightarrow{AB}$ is common and gives $(18,-16,15),(-14,16,-1)$ and generally scores **no** marks in (d).