# Question 7:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \pm((5\mathbf{i}+2\mathbf{j}+10\mathbf{k})-(2\mathbf{i}-\mathbf{j}+5\mathbf{k})) = 3\mathbf{i}+3\mathbf{j}+5\mathbf{k}$ | M1; A1 | **[2]** |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $l: \mathbf{r} = \begin{pmatrix}2\\-1\\5\end{pmatrix} + \lambda\begin{pmatrix}3\\3\\5\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}5\\2\\10\end{pmatrix} + \lambda\begin{pmatrix}3\\3\\5\end{pmatrix}$ | M1 A1ft | See notes **[2]** |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = \begin{pmatrix}-1\\1\\4\end{pmatrix} - \begin{pmatrix}2\\-1\\5\end{pmatrix} = \begin{pmatrix}-3\\2\\-1\end{pmatrix}$ | M1 | |
| $\cos\theta = \frac{\overrightarrow{AB}\bullet\overrightarrow{AD}}{|\overrightarrow{AB}||\overrightarrow{AD}|} = \frac{\begin{pmatrix}3\\3\\5\end{pmatrix}\bullet\begin{pmatrix}-3\\2\\-1\end{pmatrix}}{\sqrt{3^2+3^2+5^2}\cdot\sqrt{(-3)^2+2^2+(-1)^2}}$ | M1 | Applies dot product formula between $\overrightarrow{AB}$ (or $\overrightarrow{BA}$) and $\overrightarrow{AD}$ (or $\overrightarrow{DA}$) |
| $\cos\theta = \pm\left(\frac{-9+6-5}{\sqrt{43}\cdot\sqrt{14}}\right)$ | A1√ | Correct followed through expression |
| $\cos\theta = \frac{-8}{\sqrt{43}\cdot\sqrt{14}} \Rightarrow \theta = 109.029544... \approx 109°$ (nearest°) | A1 cso AG | awrt $109$ **[4]** |

## Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{OC} = \overrightarrow{OD} + \overrightarrow{DC} = \overrightarrow{OD} + \overrightarrow{AB} = (-\mathbf{i}+\mathbf{j}+4\mathbf{k})+(3\mathbf{i}+3\mathbf{j}+5\mathbf{k})$ | M1 | |
| $\overrightarrow{OC} = 2\mathbf{i}+4\mathbf{j}+9\mathbf{k}$ | A1 | **[2]** |

## Part (e):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $ABCD = \left(\frac{1}{2}(\sqrt{43})(\sqrt{14})\sin 109°\right) \times 2 = 23.19894905...$ | M1; dM1 A1 | awrt $23.2$ **[3]** |

## Part (f):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{\sqrt{14}} = \sin 71°$ or $\sqrt{43}\,d = 23.19894905...$ | M1 | |
| $d = \sqrt{14}\sin 71° = 3.537806563...$ | A1 | awrt $3.54$ **[2]** |

# Question 7:

## Part (a):
| $\overrightarrow{AB} = 3\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}$ | M1, A1 | Finding difference between $\overrightarrow{OB}$ and $\overrightarrow{OA}$; implied by 2/3 components correct in $3\mathbf{i}+3\mathbf{j}+5\mathbf{k}$ or $-3\mathbf{i}-3\mathbf{j}-5\mathbf{k}$ |

## Part (b):
| $\mathbf{r} = \overrightarrow{OA} + \lambda(\pm\overrightarrow{AB})$ or $\mathbf{r} = \overrightarrow{OB} + \lambda(\pm\overrightarrow{AB})$ | M1, A1ft | Expression of form (3 component vector) $\pm\lambda$(3 component vector); must write as $\mathbf{r}=$ or $l=$ or column vector form |

## Part (c):
| Attempt to find $\overrightarrow{AD}$ or $\overrightarrow{DA}$ | M1 | Implied by 2/3 components correct in $-3\mathbf{i}+2\mathbf{j}-\mathbf{k}$ or $3\mathbf{i}-2\mathbf{j}+\mathbf{k}$ |
| Apply dot product between $(\pm\overrightarrow{AB})$ and $(\pm\overrightarrow{AD})$ | M1 | |
| Angle = awrt 109° | A1ft, A1 | Dot product correctly followed through; final A1 by correct solution only. Award for dot product between $\begin{pmatrix}3\\3\\5\end{pmatrix}$ and $\begin{pmatrix}-3\\2\\-1\end{pmatrix}$ or $\begin{pmatrix}-3\\-3\\-5\end{pmatrix}$ and $\begin{pmatrix}-3\\2\\-1\end{pmatrix}$ |

**Alternative (cosine rule):**
| $|\overrightarrow{AB}|=\sqrt{43}$, $|\overrightarrow{AD}|=\sqrt{14}$, $|\overrightarrow{DB}|=\sqrt{73}$ | M1 | |
| $\cos\theta = \frac{(\sqrt{43})^2+(\sqrt{14})^2-(\sqrt{73})^2}{2\sqrt{43}\cdot\sqrt{14}}$ | M1, A1 | Cosine rule structure $\cos\theta=\frac{a^2+b^2-c^2}{2ab}$ |
| $\cos\theta = \frac{-16}{2\sqrt{43}\cdot\sqrt{14}} \Rightarrow \theta = 109°$ | A1 | awrt 109, no errors seen |

## Part (d):
| Apply $\overrightarrow{OD} +$ their $\overrightarrow{AB}$ or $\overrightarrow{OB} +$ their $\overrightarrow{AD}$ | M1 | Implied by 2/3 correctly followed through components |
| $\overrightarrow{OE} = 2\mathbf{i}+4\mathbf{j}+9\mathbf{k}$ | A1 | |

**Alternative:**
| $\overrightarrow{OE}=\begin{pmatrix}2\\-1\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\3\\5\end{pmatrix}$; set $\overrightarrow{DE}\cdot\overrightarrow{AB}=0$ to find $\lambda=-\frac{8}{43}$ | M1 | Takes dot product between $\overrightarrow{DE}$ and $\overrightarrow{AB}$, progresses to find $\lambda$ |

## Part (e):
| $\frac{1}{2}(\text{their }AB)(\text{their }CB)\sin(\text{their }109°\text{ or }71°)$ | M1 | awrt 11.6 usually implies this mark |
| Multiply by 2 for parallelogram | dM1 | |
| Area = awrt 23.2 | A1 | |

## Part (f):
| $\frac{d}{\text{their }AD}=\sin(\text{their }109°\text{ or }71°)$ or $(\text{their }AB)\cdot d = \text{their Area }ABCD$ | M1 | Award M0 if area of parallelogram = $(\text{their }AB)(\text{their }CB)$ |
| $d =$ awrt 3.54 | A1 | |

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