## Question 7:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{c} = \overrightarrow{OC} = 3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}$ | B1 cao | |
| | **[1]** | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OA} \cdot \overrightarrow{OB} = \begin{pmatrix}2\\2\\1\end{pmatrix} \cdot \begin{pmatrix}1\\1\\-4\end{pmatrix} = 2+2-4 = 0$ | M1 | An attempt to take the dot product between either $\overrightarrow{OA}$ and $\overrightarrow{OB}$, $\overrightarrow{OA}$ and $\overrightarrow{AC}$, $\overrightarrow{AC}$ and $\overrightarrow{BC}$, or $\overrightarrow{OB}$ and $\overrightarrow{BC}$ |
| $\overrightarrow{BO} \cdot \overrightarrow{BC} = \begin{pmatrix}-1\\-1\\4\end{pmatrix} \cdot \begin{pmatrix}2\\2\\1\end{pmatrix} = -2-2+4 = 0$ | A1 | Showing the result is equal to zero |
| $\overrightarrow{AC} \cdot \overrightarrow{BC} = \begin{pmatrix}1\\1\\-4\end{pmatrix} \cdot \begin{pmatrix}2\\2\\1\end{pmatrix} = 2+2-4 = 0$ | | |
| $\overrightarrow{AO} \cdot \overrightarrow{AC} = \begin{pmatrix}-2\\-2\\-1\end{pmatrix} \cdot \begin{pmatrix}1\\1\\-4\end{pmatrix} = -2-2+4 = 0$ | | |
| Therefore OA is perpendicular to OB and hence OACB is a rectangle | A1 cso | State **perpendicular** and **OACB is a rectangle** |
| Area $= 3 \times \sqrt{18} = 3\sqrt{18} = 9\sqrt{2}$ | M1 | Using distance formula to find either the correct height or width |
| | M1 | Multiplying the rectangle's height by its width |
| | A1 | exact value of $3\sqrt{18}$, $9\sqrt{2}$, $\sqrt{162}$ or aef |
| | **[6]** | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OD} = \mathbf{d} = \frac{1}{2}(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})$ | B1 | $\frac{1}{2}(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})$ |
| | **[1]** | |

### Part (d) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{DA} = \pm\left(\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{5}{2}\mathbf{k}\right)$ and $\overrightarrow{DC} = \pm\left(\frac{3}{2}\mathbf{i} + \frac{3}{2}\mathbf{j} - \frac{3}{2}\mathbf{k}\right)$ | M1 | Identifies a set of two relevant vectors |
| or $\overrightarrow{BA} = \pm(\mathbf{i} + \mathbf{j} + 5\mathbf{k})$ and $\overrightarrow{OC} = \pm(3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})$ | A1 | Correct vectors $\pm$ |
| $\cos D = (\pm)\dfrac{\begin{pmatrix}0.5\\0.5\\2.5\end{pmatrix} \cdot \begin{pmatrix}1.5\\1.5\\-1.5\end{pmatrix}}{\frac{\sqrt{27}}{2} \cdot \frac{\sqrt{27}}{2}} = (\pm)\dfrac{\frac{3}{4}+\frac{3}{4}-\frac{15}{4}}{\frac{27}{4}} = (\pm)\frac{1}{3}$ | dM1, A1$\sqrt{}$ | Applies dot product formula on multiples of these vectors. Correct ft. application of dot product formula |
| $D = \cos^{-1}\left(-\frac{1}{3}\right)$ | ddM1$\sqrt{}$ | Attempts to find the correct angle $D$ rather than $180° - D$ |
| $D = 109.47122\ldots°$ | A1 | $109.5°$ or awrt $109°$ or $1.91^c$ |
| | **[6]** | |

### Part (d) — Way 2 (Aliter):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $d\overrightarrow{BA} = \pm(\mathbf{i} + \mathbf{j} + 5\mathbf{k})$ and $d\overrightarrow{OC} = \pm(\mathbf{i} + \mathbf{j} - \mathbf{k})$ | M1 | Identifies a set of two direction vectors |
| | A1 | Correct vectors $\pm$ |
| $\cos D = (\pm)\dfrac{\begin{pmatrix}1\\1\\-1\end{pmatrix} \cdot \begin{pmatrix}1\\1\\5\end{pmatrix}}{\sqrt{3}\cdot\sqrt{27}} = (\pm)\dfrac{1+1-5}{\sqrt{3}\cdot\sqrt{27}} = (\pm)\frac{1}{3}$ | dM1 | Applies dot product formula on multiples of these vectors |
| | A1$\sqrt{}$ | Correct ft. application of dot product formula |
| $D = \cos^{-1}\left(-\frac{1}{3}\right)$ | ddM1$\sqrt{}$ | Attempts to find the correct angle $D$ rather than $180° - D$ |
| $D = 109.47122\ldots°$ | A1 | $109.5°$ or awrt $109°$ or $1.91^c$ |
| | **[6]** | |

## Question (d) – Aliter Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $d\overrightarrow{OA} = (2\mathbf{i} + 2\mathbf{j} + \mathbf{k})$ and $d\overrightarrow{OC} = (\mathbf{i} + \mathbf{j} - \mathbf{k})$ | M1, A1 | Identifies a set of two direction vectors; Correct vectors |
| $\cos(\frac{1}{2}D) = \frac{\begin{pmatrix}2\\2\\1\end{pmatrix}\cdot\begin{pmatrix}1\\1\\-1\end{pmatrix}}{\sqrt{9}\cdot\sqrt{3}} = \frac{2+2-1}{\sqrt{9}\cdot\sqrt{3}} = \frac{1}{\sqrt{3}}$ | dM1, A1$\checkmark$ | Applies dot product formula on multiples of these vectors; Correct ft application of dot product formula |
| $D = 2\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$ | ddM1$\checkmark$ | Attempts to find correct angle D by doubling their angle for $\frac{1}{2}D$ |
| $D = 109.47122...°$ | A1 | $109.5°$ or awrt $109°$ or $1.91^c$ |

**[6]**

---

## Question (d) – Aliter Way 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{DA} = \frac{1}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}+\frac{5}{2}\mathbf{k}$, $\overrightarrow{DC} = \frac{3}{2}\mathbf{i}+\frac{3}{2}\mathbf{j}-\frac{3}{2}\mathbf{k}$, $\overrightarrow{AC} = \mathbf{i}+\mathbf{j}-4\mathbf{k}$ | — | Using cosine rule |
| $|\overrightarrow{DA}| = \frac{\sqrt{27}}{2}$, $|\overrightarrow{DC}| = \frac{\sqrt{27}}{2}$, $|\overrightarrow{AC}| = \sqrt{18}$ | M1, A1 | Attempts to find all lengths of all three edges of $\triangle ADC$; All Correct |
| $\cos D = \frac{\left(\frac{\sqrt{27}}{2}\right)^2 + \left(\frac{\sqrt{27}}{2}\right)^2 - \left(\sqrt{18}\right)^2}{2\left(\frac{\sqrt{27}}{2}\right)\left(\frac{\sqrt{27}}{2}\right)} = -\frac{1}{3}$ | dM1, A1$\checkmark$ | Using cosine rule formula with correct 'subtraction'; Correct ft application of cosine rule formula |
| $D = \cos^{-1}\left(-\frac{1}{3}\right)$ | ddM1$\checkmark$ | Attempts to find correct angle D rather than $180° - D$ |
| $D = 109.47122...°$ | A1 | $109.5°$ or awrt $109°$ or $1.91^c$ |

**[6]**

---

## Question (d) – Aliter Way 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{DA} = \frac{1}{2}\mathbf{i}+\frac{1}{2}\mathbf{j}+\frac{5}{2}\mathbf{k}$, $\overrightarrow{OA} = 2\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $\overrightarrow{AC} = \mathbf{i}+\mathbf{j}-4\mathbf{k}$; Let X be midpoint of AC | — | Using trigonometry on a right angled triangle |
| $|\overrightarrow{DA}| = \frac{\sqrt{27}}{2}$, $|\overrightarrow{DX}| = \frac{1}{2}|\overrightarrow{OA}| = \frac{3}{2}$, $|\overrightarrow{AX}| = \frac{1}{2}|\overrightarrow{AC}| = \frac{1}{2}\sqrt{18}$ | M1, A1 | Attempts to find two out of three lengths in $\triangle ADX$; Any two correct |
| $\sin(\frac{1}{2}D) = \frac{\frac{\sqrt{18}}{2}}{\frac{\sqrt{27}}{2}}$, $\cos(\frac{1}{2}D) = \frac{\frac{3}{2}}{\frac{\sqrt{27}}{2}}$ or $\tan(\frac{1}{2}D) = \frac{\frac{\sqrt{18}}{2}}{\frac{3}{2}}$ | dM1, A1$\checkmark$ | Uses correct sohcahtoa to find $\frac{1}{2}D$; Correct ft application of sohcahtoa |
| $D = 2\tan^{-1}\left(\frac{\frac{\sqrt{18}}{2}}{\frac{3}{2}}\right)$ | ddM1$\checkmark$ | Attempts to find correct angle D by doubling their angle for $\frac{1}{2}D$ |
| $D = 109.47122...°$ | A1 | $109.5°$ or awrt $109°$ or $1.91^c$ |

**[6]**

---

## Question (d) – Aliter Way 6:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OC} = 3\mathbf{i}+3\mathbf{j}-3\mathbf{k}$, $\overrightarrow{OA} = 2\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $\overrightarrow{AC} = \mathbf{i}+\mathbf{j}-4\mathbf{k}$ | — | Using trigonometry on a right angled similar triangle OAC |
| $|\overrightarrow{OC}| = \sqrt{27}$, $|\overrightarrow{OA}| = 3$, $|\overrightarrow{AC}| = \sqrt{18}$ | M1, A1 | Attempts to find two out of three lengths in $\triangle OAC$; Any two correct |
| $\sin(\frac{1}{2}D) = \frac{\sqrt{18}}{\sqrt{27}}$, $\cos(\frac{1}{2}D) = \frac{3}{\sqrt{27}}$ or $\tan(\frac{1}{2}D) = \frac{\sqrt{18}}{3}$ | dM1, A1$\checkmark$ | Uses correct sohcahtoa to find $\frac{1}{2}D$; Correct ft application of sohcahtoa |
| $D = 2\tan^{-1}\left(\frac{\sqrt{18}}{3}\right)$ | ddM1$\checkmark$ | Attempts to find correct angle D by doubling their angle for $\frac{1}{2}D$ |
| $D = 109.47122...°$ | A1 | $109.5°$ or awrt $109°$ or $1.91^c$ |

**[6]**

---

## Question 7(b)(i) – Aliter Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{c} = \overrightarrow{OC} = \pm(3\mathbf{i}+3\mathbf{j}-3\mathbf{k})$, $\overrightarrow{AB} = \pm(-\mathbf{i}-\mathbf{j}-5\mathbf{k})$ | — | — |
| $|\overrightarrow{OC}| = \sqrt{(3)^2+(3)^2+(-3)^2} = \sqrt{(1)^2+(1)^2+(-5)^2} = |\overrightarrow{AB}|$ | M1 | A complete method of proving that the diagonals are equal |
| $|\overrightarrow{OC}| = |\overrightarrow{AB}| = \sqrt{27}$ | A1 | Correct result |
| Diagonals are equal, and OACB is a rectangle | A1 cso | Diagonals are equal and OACB is a rectangle |

**[3]**

---

## Question 7(b)(i) – Aliter Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(OA)^2 + (AC)^2 = (OC)^2$ or $(BC)^2 + (OB)^2 = (OC)^2$ or $(OA)^2 + (OB)^2 = (AB)^2$ or $(BC)^2 + (AC)^2 = (AB)^2$ | — | — |
| $(3)^2 + (\sqrt{18})^2 = (\sqrt{27})^2$ | M1, A1 | A complete method of proving Pythagoras holds using their values; Correct result |
| OA is perpendicular to OB, or AC is perpendicular to BC, and hence OACB is a rectangle | A1 cso | Perpendicular and OACB is a rectangle |

**[3] [14 marks total]**

---