# Question 8:

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $A(3,5,0)$ | B1 | $(3,5,0)$ |

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\{l_2:\}\,\mathbf{r}=\begin{pmatrix}1\\5\\2\end{pmatrix}+\lambda\begin{pmatrix}-5\\4\\3\end{pmatrix}$ | M1 | $\mathbf{a}+\lambda\mathbf{d}$ or $\mathbf{a}+\mu\mathbf{d}$, $\mathbf{a}+t\mathbf{d}$, $\mathbf{a}\neq\mathbf{0}$, $\mathbf{d}\neq\mathbf{0}$, with either $\mathbf{a}=\mathbf{i}+5\mathbf{j}+2\mathbf{k}$ or $\mathbf{d}=-5\mathbf{i}+4\mathbf{j}+3\mathbf{k}$, or a multiple of $-5\mathbf{i}+4\mathbf{j}+3\mathbf{k}$ |
| Correct vector equation using $\mathbf{r}=$ or $l=$ or $l_2=$ | A1 | Do not allow $l_2$: or $l_2\rightarrow$ or $l_1=$ for the A1 mark |

## Part (c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{AP}=\overrightarrow{OP}-\overrightarrow{OA}=\begin{pmatrix}1\\5\\2\end{pmatrix}-\begin{pmatrix}3\\5\\0\end{pmatrix}=\begin{pmatrix}-2\\0\\2\end{pmatrix}$ | M1 | Full method for finding $AP$ |
| $AP=\sqrt{(-2)^2+(0)^2+(2)^2}=\sqrt{8}=2\sqrt{2}$ | A1 | $2\sqrt{2}$ |

## Part (d):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{AP}\cdot\mathbf{d}_2$ required | M1 | Realisation that the dot product is required between $(\overrightarrow{AP}$ or $\overrightarrow{PA})$ and $\pm K\mathbf{d}_2$ or $\pm K\mathbf{d}_1$ |
| $\cos\theta=\frac{\overrightarrow{AP}\cdot\mathbf{d}_2}{|\overrightarrow{AP}||\mathbf{d}_2|}=\frac{\pm\left(\begin{pmatrix}-2\\0\\2\end{pmatrix}\cdot\begin{pmatrix}-5\\4\\3\end{pmatrix}\right)}{\sqrt{(-2)^2+(0)^2+(2)^2}\cdot\sqrt{(-5)^2+(4)^2+(3)^2}}$ | dM1 | Dependent on previous M mark. Applies dot product formula between their $(\overrightarrow{AP}$ or $\overrightarrow{PA})$ and $\pm K\mathbf{d}_2$ or $K\mathbf{d}_1$ |
| $\cos\theta=\frac{\pm(10+0+6)}{\sqrt{8}\cdot\sqrt{50}}=\frac{4}{5}$ | A1 cso | $\cos\theta=\frac{4}{5}$ or $0.8$ or $\frac{8}{10}$ or $\frac{16}{20}$ |

## Part (e):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Area }APE=\frac{1}{2}(\text{their }2\sqrt{2})^2\sin\theta$ | M1 | $\frac{1}{2}(\text{their }2\sqrt{2})^2\sin\theta$ or $\frac{1}{2}(\text{their }2\sqrt{2})^2\sin(\text{their }\theta)$ |
| $=2.4$ | A1 | $2.4$ or $\frac{12}{5}$ or $\frac{24}{10}$ or awrt $2.40$ |

## Part (f):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{PE}=(-5\lambda)\mathbf{i}+(4\lambda)\mathbf{j}+(3\lambda)\mathbf{k}$ and $PE=\text{their }2\sqrt{2}$ from part (c) | — | — |
| $(-5\lambda)^2+(4\lambda)^2+(3\lambda)^2=(\text{their }2\sqrt{2})^2$ | M1 | This mark can be implied |
| $50\lambda^2=8\Rightarrow\lambda^2=\frac{4}{25}\Rightarrow\lambda=\pm\frac{2}{5}$ | A1 | Either $\lambda=\frac{2}{5}$ or $\lambda=-\frac{2}{5}$ |
| Substitutes at least one value of $\lambda$ into $l_2$ | dM1 | Dependent on previous M mark |
| $\{\overrightarrow{OE}\}=\begin{pmatrix}3\\\frac{17}{5}\\4\end{pmatrix}$ or $\begin{pmatrix}3\\3.4\\0.8\end{pmatrix}$, $\{\overrightarrow{OE}\}=\begin{pmatrix}-1\\\frac{33}{5}\\\frac{16}{5}\end{pmatrix}$ or $\begin{pmatrix}-1\\6.6\\3.2\end{pmatrix}$ | A1 | At least one set of coordinates correct |
| Both sets of coordinates correct | A1 | — |

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