# Question 1:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\3\\-2\end{pmatrix} \times \begin{pmatrix}-3\\-6\\4\end{pmatrix}$ | M1 | Use of cross product. Correct pairs of numbers used together, allow one numerical or sign error. |
| $\begin{pmatrix}0\\2\\3\end{pmatrix}$ | A1 | Or any non-zero multiple |

**Alt method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3-2a=0$, assuming vector of form $\begin{pmatrix}0\\1\\a\end{pmatrix}$, both dot products zero | M1 | Allow one numerical or sign error |
| $\begin{pmatrix}0\\1\\1.5\end{pmatrix}$ | A1 | Or any non-zero multiple |

**Alt method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a+3b-2c=0$, $-3a-6b+4c=0$, $\Rightarrow a=0, 3b-2c=0$ | M1 | Use dot product with both vectors, eliminate $a$ or $b$ & $c$ |
| $\begin{pmatrix}0\\2\\3\end{pmatrix}$ | A1 | Or any non-zero multiple |

**[2 marks total]**

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## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{x-0}{2}$ or $\frac{y-3}{1}$ or $\frac{z-(-2)}{\frac{1}{2}}$ | M1 | Or $x=(0+)2\lambda$, $y=3+(1)\lambda$, $z=\frac{\lambda-4}{2}$, or form $(\mathbf{r}-\mathbf{a})\times\mathbf{b}=\mathbf{0}$. Can be implied by correct answer. |
| $\mathbf{r}=\begin{pmatrix}0\\3\\-2\end{pmatrix}+\ldots$ | A1 | Could have any multiple of direction vector added, e.g. $[4,5,-1]$. If M0 then SC1 for $\begin{pmatrix}0\\3\\-2\end{pmatrix}+\ldots$ (no $\lambda$) |
| $\ldots+\lambda\begin{pmatrix}2\\1\\\frac{1}{2}\end{pmatrix}$ | A1 | Vector could be any multiple e.g. $[4,2,1]$. Any sensible parameter name (not e.g. $r,x,y$ or $z$). Not dependent on previous A mark, M1 A0 A1 possible. |

**[3 marks total]**

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