# Question 4:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds any two vectors $\pm\overrightarrow{LM}$, $\pm\overrightarrow{LN}$ or $\pm\overrightarrow{MN}$: e.g. $\pm\begin{pmatrix}8\\1\\1\end{pmatrix}$ or $\pm\begin{pmatrix}4\\3\\1\end{pmatrix}$ or $\pm\begin{pmatrix}-4\\2\\0\end{pmatrix}$; two out of three correct is sufficient | M1 | Implies correct method |
| Applies $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ where $\mathbf{a}$ is any coordinate from L, M or N | M1 | Vectors $\mathbf{b}$ and $\mathbf{c}$ from attempt at two vectors lying on plane |
| Correct equation: $\mathbf{a}=\begin{pmatrix}-2\\-3\\-1\end{pmatrix}$ or $\begin{pmatrix}6\\-2\\0\end{pmatrix}$ or $\begin{pmatrix}2\\0\\0\end{pmatrix}$; $\mathbf{b}$ and $\mathbf{c}$ any two from the above vectors | A1 | |

## Parts (b)(i) and (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Applies $\mathbf{b}\cdot\begin{pmatrix}1\\2\\-10\end{pmatrix}$ AND $\mathbf{c}\cdot\begin{pmatrix}1\\2\\-10\end{pmatrix}$ | M1 | AO 1.1b |
| Shows both dot products $= 0$, therefore lawn is **perpendicular** | A1 | AO 2.4 |
| **Alt 1:** Shows result is parallel to $\begin{pmatrix}1\\2\\-10\end{pmatrix}$, therefore lawn is **perpendicular** | A1 | |
| **Alt 2:** Achieves constant value 2; concludes as constant therefore lawn is **perpendicular** | A1 | |
| **Outside Spec:** Finds cross product between $\mathbf{b}$ and $\mathbf{c}$ | M1 | AO 1.1b |

# Question 4 (Vectors/Lawn Problem):

## Part (b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Compares vectors **b** and **c** with $\begin{pmatrix}1\\2\\-10\end{pmatrix}$ to show parallel, OR applies dot product formula with $\begin{pmatrix}1\\2\\-10\end{pmatrix}$ | M1 | Two of three values correct sufficient |
| Concludes **parallel** therefore lawn is **perpendicular** | A1 | 2.4 |

## Part (b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-10\end{pmatrix} = \mathbf{a}\cdot\begin{pmatrix}1\\2\\-10\end{pmatrix}$ where $\mathbf{a}=\begin{pmatrix}-2\\-3\\-1\end{pmatrix}$ or $\begin{pmatrix}6\\-3\\0\end{pmatrix}$ or $\begin{pmatrix}2\\0\\0\end{pmatrix}$ | M1 | 1.1b; Allow $\mathbf{r}\cdot\begin{pmatrix}1\\2\\-10\end{pmatrix}=\mathbf{a}\cdot\begin{pmatrix}1\\2\\-10\end{pmatrix}$ |
| $x+2y-10z=2$ or $x+2y-10z-2=0$ | A1 | 1.1b |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds vector $\overrightarrow{PQ}$ or $\overrightarrow{QP}$ and uses as direction vector in $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ | M1 | 3.3; Two out of three values correct sufficient |
| $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ where $\mathbf{a}=\begin{pmatrix}-10\\8\\2\end{pmatrix}$ or $\begin{pmatrix}6\\4\\3\end{pmatrix}$ and $\mathbf{d}=\pm\begin{pmatrix}16\\-4\\1\end{pmatrix}$ | A1 | 1.1b |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. "The lawn will not be flat" or "The washing line will not be straight" | B1 | 3.5b |

## Part (e):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{\lvert(2\times1)+5\times2+(2.75\times-10)-2\rvert}{\sqrt{1^2+2^2+(-10)^2}}$ | M1 | 3.4 |
| $=1.71$ m or $171$ cm | A1 | 2.2b |

## Part (f):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Compares answer to (e) with $1.5$ m and makes consistent assessment of model | B1ft | 3.5a; If answer close to $1.5$ (e.g. $1.4$–$1.6$) must conclude model is good; if significantly different must conclude model is not good. No contradictory statements. |

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