# Question 8:

## Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| If $C$ is not at $A$ or $B$: angle $ACB$ is "an angle in a semi-circle" so $90°$ | **B1** | Could choose to prove the circle theorem. |
| $\overrightarrow{AC} \cdot \overrightarrow{BC} = 0$ since $\overrightarrow{AC}$ and $\overrightarrow{BC}$ are perpendicular and the dot product of perpendicular vectors is zero (because $\cos 90° = 0$) | **B1** | A reason must be given. Just "$\overrightarrow{AC} \cdot \overrightarrow{BC} = 0$" is insufficient. The dot product of perpendicular vectors is zero. Implication must be the correct way around. |
| In the case where $C = A$ or $C = B$ then $\overrightarrow{AC} = \mathbf{0}$ or $\overrightarrow{BC} = \mathbf{0}$ and so $\overrightarrow{AC} \cdot \overrightarrow{BC} = 0$ since the zero vector dotted with any vector is zero. | **B1** | Must mention both cases but second case can just be covered by "Similarly for $B$" or "eg consider $A$". |

**Alternative proof (coordinate method):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let centre of sphere be at $O$. Let $A = (\alpha, \beta, \gamma)$ where $\alpha^2 + \beta^2 + \gamma^2 = r^2$. Since $AB$ is a diameter, $B = (-\alpha, -\beta, -\gamma)$. Let $C = (x,y,z)$ where $x^2+y^2+z^2=r^2$. $\overrightarrow{AC} = \begin{pmatrix}x-\alpha\\y-\beta\\z-\gamma\end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix}x+\alpha\\y+\beta\\z+\gamma\end{pmatrix}$ | **B1** | Centre at $O$ and attempting $\overrightarrow{AC}$ or $\overrightarrow{BC}$ |
| $\overrightarrow{AC} \cdot \overrightarrow{BC} = (x-\alpha)(x+\alpha)+(y-\beta)(y+\beta)+(z-\gamma)(z+\gamma)$ | **B1** | $\overrightarrow{AC}$ and $\overrightarrow{BC}$ both correct and dot product formed |
| $\overrightarrow{AC} \cdot \overrightarrow{BC} = (x^2+y^2+z^2)-(\alpha^2+\beta^2+\gamma^2) = r^2 - r^2 = 0$ | **B1** | Correctly showing that dot product is $0$ |

**Alternative proof (vector method):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let position vectors of $A$, $B$, $C$ be $\mathbf{a}, \mathbf{b}, \mathbf{c}$. We have $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=r$. $\overrightarrow{AC}\cdot\overrightarrow{BC} = (\mathbf{c}-\mathbf{a})\cdot(\mathbf{c}-\mathbf{b}) = |\mathbf{c}|^2 - \mathbf{a}\cdot\mathbf{c} - \mathbf{b}\cdot\mathbf{c} + \mathbf{a}\cdot\mathbf{b}$ | **B1** | Centre at $O$ and attempting $\overrightarrow{AC}$ or $\overrightarrow{BC}$ |
| $= |\mathbf{c}|^2 - (\mathbf{a}+\mathbf{b})\cdot\mathbf{c} + \mathbf{a}\cdot\mathbf{b}$. But $\mathbf{a} = -\mathbf{b}$ (as $A$, $B$ ends of diameter): $\overrightarrow{AC}\cdot\overrightarrow{BC} = |\mathbf{c}|^2 - (\mathbf{a}-\mathbf{a})\cdot\mathbf{c} - \mathbf{a}\cdot\mathbf{a} = |\mathbf{c}|^2 - |\mathbf{a}|^2$ | **B1** | $\overrightarrow{AC}$ and $\overrightarrow{BC}$ both correct and dot product formed |
| $= r^2 - r^2 = 0$ | **B1** | Stating and using $\mathbf{a} = -\mathbf{b}$ to show that dot product is $0$ |

**[3 marks total]**

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## Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AC} = \begin{pmatrix}2p-11\\p-12\\1--14\end{pmatrix}$ or $\overrightarrow{BC} = \begin{pmatrix}2p-9\\p-13\\1-6\end{pmatrix}$ | **M1** | Attempt to subtract (in either order) |
| $\overrightarrow{AC} = \begin{pmatrix}2p-11\\p-12\\15\end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix}2p-9\\p-13\\-5\end{pmatrix}$ | **A1** | Both completely correct (could be $\overrightarrow{CA} = \begin{pmatrix}11-2p\\12-p\\-15\end{pmatrix}$ etc) |
| $\overrightarrow{AC}\cdot\overrightarrow{BC} = 0 \Rightarrow \begin{pmatrix}2p-11\\p-12\\15\end{pmatrix}\cdot\begin{pmatrix}2p-9\\p-13\\-5\end{pmatrix} = $ | **M1** | Attempt to dot their $\overrightarrow{AC}$ and $\overrightarrow{BC}$. Dot product must result in a scalar quantity. "= 0" not necessary for M1. |
| $(2p-11)(2p-9)+(p-12)(p-13)+15(-5)=0$ $5p^2-65p+180=0$ or $p^2-13p+36=0$ | **A1** | Correct three term quadratic (must have "= 0" appearing somewhere, might be near start) |
| $p=4$ or $p=9$ | **A1** | |
| So the possible locations are $(8,4,1)$ or $(18,9,1)$ | **A1FT** | Solution must be given as coordinates. For FT must have solved a quadratic coming from attempt at dot product. |

**Alternative method 1 (Pythagoras):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $|AB|^2 = 2^2+1^2+20^2 = 405$; $|AC|^2=(11-2p)^2+(12-p)^2+(-14-1)^2$; $|BC|^2=(9-2p)^2+(13-p)^2+(6-1)^2$ | **M1, A1** | Attempting to find $|AC|^2$ or $|BC|^2$. Correct expressions for $|AC|^2$ and $|BC|^2$. |
| $|AB|^2 = |AC|^2+|BC|^2$; $10p^2-130p+765=405$ | **M1, A1** | Using Pythagoras. Correct 3-term quadratic. |
| $p^2-13p+36=0$; $p=4$ or $p=9$ | **A1** | |
| Possible locations are $(8,4,1)$ or $(18,9,1)$ | **A1FT** | Solution must be given as coordinates |

**Alternative method 2 (Centre of sphere):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre of sphere at $(10, 12.5, -4)$; $r^2=(11-10)^2+(12-12.5)^2+(-14+4)^2 = \frac{405}{4}$ | **M1** | Attempting to find $r^2$ (could be via diameter) |
| $|OC|^2=(2p-10)^2+(p-12.5)^2+(1+4)^2 = \frac{405}{4}$ | **M1, A1** | Attempting to find $|OC|^2$. Correct expression. |
| $p^2-13p+36=0$; $p=4$ or $p=9$ | **A1, A1** | Correct 3-term quadratic |
| Possible locations are $(8,4,1)$ or $(18,9,1)$ | **A1FT** | Solution must be given as coordinates |

**[6 marks total]**