## Question 9(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-1\times-2+2\times3+1\times k=0$ | M1 | Attempt scalar product and set equal to zero; allow use of **i, j, k** notation |
| $\Rightarrow k=-4$ | A1 | |

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## Question 9(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate $x$ and $y$ coordinates: $3+\lambda=6+2\mu\Rightarrow\lambda-2\mu=3$; $2-\lambda=5+\mu\Rightarrow\lambda+\mu=-3$ | M1 | Use coordinates to find $\mu$ and $\lambda$ |
| $\Rightarrow\mu=-2,\lambda=-1$ | A1 | |
| Consistent with $z$ coordinates since $7+3\times(-1)=4$ and $2-(-2)=4$ | E1 | Check consistency with third coordinate |
| So the point of intersection is $(2,3,4)$ | A1 | |

## Question 9(iii):

**Method 1 (Vector Product):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1\\-1\\3\end{pmatrix} \times \begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}-2\\-7\\3\end{pmatrix}$ | M1, A1 | Attempt the vector product by any valid method; BC |
| $\begin{pmatrix}1\\a\\b\end{pmatrix} = \lambda\begin{pmatrix}-2\\-7\\3\end{pmatrix}$ | M1 | |
| $\lambda = -\frac{1}{2}$ | M1 | |
| $a = 3.5,\ b = -1.5$ | A1 [5] | |

**Method 2 (Dot Products):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1\\-1\\3\end{pmatrix}\cdot\begin{pmatrix}1\\a\\b\end{pmatrix} = 0$ and $\begin{pmatrix}2\\1\\-1\end{pmatrix}\cdot\begin{pmatrix}1\\a\\b\end{pmatrix} = 0$ | M1A1 | |
| $1 - a + 3b = 0$ and $2 + a - b = 0$ | A1 | |
| Solve simultaneous equations | M1 | |
| $a = 3.5,\ b = -1.5$ | A1 | |