# Question 1:

## Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$-coord (or $y$-coord): $4 + 3\lambda = 19$ or $-2 + -2\lambda = -12 \Rightarrow \lambda = 5$ | M1 | Forming and solving an equation in $\lambda$ for any coordinate |
| $z$-coord: $7 + 4\lambda = 17 \Rightarrow \lambda = 2.5$ (or if $\lambda = 5$ then $7 + 4\lambda = 27$) (or $7 + 4 \times 5 \neq 17$) | M1 | Either forming and solving an equation with a different correct solution in $\lambda$ or using the previous value to demonstrate an inconsistency. This second M mark is for considering a coordinate with a different $\lambda$ (So M2 for $z$ and one other) |
| Inconsistency so point does **not** lie on the line | A1 | Correct conclusion e.g. "point not on line is enough here" as long as with 2 correct different values of $\lambda$ – no need to see the word "inconsistency" |
| **[3]** | | |

## Part (b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{a.b} = 1\times-3 + -2\times6 + 2\times2$ | M1 | Forming the dot product. Can be implied by $-11$ (but not 11) |
| $\cos\theta = \dfrac{"-11"}{\sqrt{1^2+(-2)^2+2^2}\sqrt{(-3)^2+6^2+2^2}}$ | M1 | Their dot product divided by the product of the (correctly formed) moduli. Allow sin/cos confusion here. $-\dfrac{11}{21}$ but can be awarded for $\dfrac{11}{21}$ |
| $\theta = 122°$ (3 sf) | A1 | Do not ISW (so e.g. M2A0 for working leading to final answer of $58.4°$). $121.5881…$ If more accurate than 3 s.f. accept answers in range $[121.4, 121.8]$ |
| **[3]** | | |

## Part (b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Required vector is $\mathbf{a}\times\mathbf{b}$ | M1 | This mark can be awarded for any non-zero multiple, even if non-numerical. Some evidence of method needed for M1. Could be one correct value, or correct calculation for one value. |
| $=\begin{pmatrix}1\\-2\\2\end{pmatrix}\times\begin{pmatrix}-3\\6\\2\end{pmatrix}=\begin{pmatrix}-16\\-8\\0\end{pmatrix}$ | A1 | or any numerical non-zero multiple e.g. $\begin{pmatrix}2\\1\\0\end{pmatrix}$. Ignore errors in attempts to simplify. Condone incorrect statements e.g. $\begin{pmatrix}-16\\-8\\0\end{pmatrix}=\begin{pmatrix}2\\1\\0\end{pmatrix}$ |
| **Alternative:** Assume vector of form $\begin{pmatrix}1\\p\\q\end{pmatrix}$, then $1-2p+2q=0$ and $-3+6p+2q=0$ | M1 | Need both 'dot product' equations. Ignore lack of consideration of $\begin{pmatrix}0\\p\\q\end{pmatrix}$. Could also be awarded for considering $\begin{pmatrix}t\\p\\q\end{pmatrix}$ and deriving $t-2p+2q=0$ and $-3t+6p+2q=0$ |
| $2p-2q=1$ and $6p+2q=3 \Rightarrow p=\frac{1}{2}, q=0$, so required vector is $\begin{pmatrix}1\\\frac{1}{2}\\0\end{pmatrix}$ | A1 | or any numerical non-zero multiple e.g. $\begin{pmatrix}2\\1\\0\end{pmatrix}$. Eliminating e.g. $q$ leads to $t=2p$ and then $q=0$ and so to $\begin{pmatrix}2p\\p\\0\end{pmatrix}$ but a (non-zero) value for $p$ must be chosen so that the final answer is a vector and not a family of vectors. |
| **[2]** | | |

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