## Question 8(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct direction vector seen or implied ($\overrightarrow{BC} = 3\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}$) | B1 | Condone $\overrightarrow{BC} = -3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}$. |
| Use a correct method to form a vector equation | M1 | Allow for the RHS with no LHS. |
| Obtain $\mathbf{r} = 5\mathbf{i} + 2\mathbf{j} + \lambda(\mathbf{i} + \mathbf{j} - \mathbf{k})$ | A1 | ISW. Must have $\mathbf{r} = ...$ or $\begin{pmatrix}x\\y\\z\end{pmatrix} = ...$, not $l_1 = ...$. Or equivalent vector form, e.g. $\mathbf{r} = 8\mathbf{i}+5\mathbf{j}-3\mathbf{k}+\alpha(\mathbf{i}+\mathbf{j}-\mathbf{k})$ or $\mathbf{r} = 5\mathbf{i}+2\mathbf{j}+\lambda(3\mathbf{i}+3\mathbf{j}-3\mathbf{k})$. Condone a column vector with $\mathbf{i}, \mathbf{j}, \mathbf{k}$. |
| | **3** | |

## Question 8(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use components to form two relevant equations in 2 unknowns. For their $l_1$ B0 if they use the same unknown for both lines. | B1FT | Two components of $\begin{pmatrix}5+\lambda\\2+\lambda\\-\lambda\end{pmatrix} = \begin{pmatrix}-2+3\mu\\1+\mu\\4-2\mu\end{pmatrix}$ seen or implied. |
| Solve 2 relevant equations in 2 unknowns for $\lambda$ or $\mu$ | M1 | For *their* $l_1$. |
| Obtain $\lambda = 2$ or $\mu = 3$ | A1 | Or equivalent e.g. using $\overrightarrow{BC}$ as direction vector gives $\lambda = \frac{2}{3}$. |
| Obtain $(7, 4, -2)$. No need to check the third equation – the question implies that the lines intersect. | A1 | Accept position vector. Condone a column vector with $\mathbf{i}, \mathbf{j}, \mathbf{k}$. **SC: B1 M1 A1 A1** if one component of their line is incorrect but they do not use that component. |
| | **4** | |

## Question 8(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $AB = \sqrt{7^2 + 1^2 + 4^2} \left(= \sqrt{66}\right)$ | B1 | Or $(AB)^2 = 66$. Condone a sign error in $\overrightarrow{AB}$ |
| State $\overrightarrow{BD}$ in component form: $\begin{pmatrix} -7+3r \\ -1+r \\ 4-2r \end{pmatrix}$ | B1 | Or equivalent |
| $AB = BD \Rightarrow (3r-7)^2 + (r-1)^2 + (-2r+4)^2 = 66$, giving $\left(14r^2 - 60r = 0\right)$ | M1 | Or equivalent equation in one unknown for their $AB$ and their $\overrightarrow{BD} \neq \overrightarrow{OD}$. If you never see a correct form and they go direct to $9r^2 + 49 + r^2 + 1...$ then M0 |
| $\Rightarrow r = \frac{30}{7}$ | A1 | Correct only. Ignore $r = 0$ if seen |
| $\overrightarrow{OD} = \frac{76}{7}\mathbf{i} + \frac{37}{7}\mathbf{j} - \frac{32}{7}\mathbf{k}$ | A1 | Must be a vector. Condone if also have $\overrightarrow{OD} = \overrightarrow{OA}$ |

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