## Question 7:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x-3}{4}=\frac{y-5}{-2}=\frac{z-4}{7} \Rightarrow \mathbf{r}=\begin{pmatrix}3\\5\\4\end{pmatrix}\pm\lambda\begin{pmatrix}4\\-2\\7\end{pmatrix}$ | M1 | Converts to parametric form. "$\mathbf{r}=$" is not required |
| $2x+4y-z=1 \Rightarrow 2(3+4\lambda)+4(5-2\lambda)-4-7\lambda=1 \Rightarrow \lambda=\ldots(3) \Rightarrow P$ is... | M1 | Correct strategy for finding $P$. Condone use of $2x+4y-z=0$ for the plane equation |
| $(15,-1,25)$ | A1 | Correct coordinates. Condone if given as a vector |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x-3}{4}=\frac{y-5}{-2} \Rightarrow x=13-2y$ | M1 | Uses the Cartesian equation to find $x$ in terms of $y$ |
| $2x+4y-z=1 \Rightarrow 26-4y+4y-z=1 \Rightarrow z=\ldots, x=\ldots, y=\ldots$ | M1 | Correct strategy for finding $P$. Condone use of $2x+4y-z=0$ |
| $(15,-1,25)$ | A1 | Correct coordinates. Condone if given as a vector |

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### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}4\\-2\\7\end{pmatrix}\bullet\begin{pmatrix}2\\4\\-1\end{pmatrix}=8-8-7=-7$ | M1 | Applies the scalar product between the direction of $l_1$ and the normal to the plane |
| $\phi=\cos^{-1}\frac{\pm7}{\sqrt{69}\sqrt{21}}=\ldots$ or $\phi=\sin^{-1}\frac{\pm7}{\sqrt{69}\sqrt{21}}=\ldots$ | dM1 | Attempts to find a relevant angle in degrees or radians. Depends on the first method mark |
| $\theta=10.6°$ | A1 | Allow awrt 10.6 but do **not** isw. For reference $\theta=10.5965654°$ |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}4\\-2\\7\end{pmatrix}\times\begin{pmatrix}2\\4\\-1\end{pmatrix}=\begin{pmatrix}26\\-18\\-20\end{pmatrix}$ | M1 | Attempts vector product of normal to $\Pi$ and direction of $l_1$ |
| $\sqrt{26^2+18^2+20^2}=\sqrt{21}\sqrt{69}\sin\alpha$, $\sin\alpha=\frac{10\sqrt{46}}{69}\Rightarrow\alpha=\ldots$ | dM1 | Attempts to find a relevant angle. Depends on the first method mark |
| $\theta=10.6°$ | A1 | Allow awrt 10.6 but do **not** isw. For reference $\theta=10.5965654°$ |

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### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&4&-1\\4&-2&7\end{vmatrix}=\begin{pmatrix}26\\-18\\-20\end{pmatrix}$ | M1 | Attempts vector product of normal to $\Pi$ and direction of $l_1$. If no method is seen expect at least 2 correct components |
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\13&-9&-10\\2&4&-1\end{vmatrix}=\begin{pmatrix}49\\-7\\70\end{pmatrix}$ | M1 | Attempts vector product of "$\mathbf{a}$" with normal to $\Pi$ to find direction of $l_2$ |
| Correct direction for $l_2$ | A1 | Correct direction for $l_2$ |
| $\mathbf{r}=\begin{pmatrix}15\\-1\\25\end{pmatrix}+\mu\begin{pmatrix}7\\-1\\10\end{pmatrix}$ | ddM1 | Depends on both previous M marks. Attempts vector equation using their direction vector and their $P$ |
| Correct equation or any equivalent correct vector equation | A1 | |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda=1 \Rightarrow (7,3,11)$ lies on $l_1$; $\mathbf{r}=\begin{pmatrix}7\\3\\11\end{pmatrix}+t\begin{pmatrix}2\\4\\-1\end{pmatrix}$; $\Rightarrow 2(7+2t)+4(3+4t)-11+t=1$; $t=-\frac{2}{3}\Rightarrow\left(\frac{17}{3},\frac{1}{3},\frac{35}{3}\right)$ is on $l_2$ | M1 | Complete method to find a point on $l_2$ |
| Direction of $l_2$ is $\begin{pmatrix}15\\-1\\25\end{pmatrix}-\frac{1}{3}\begin{pmatrix}17\\1\\35\end{pmatrix}=\frac{1}{3}\begin{pmatrix}28\\-4\\40\end{pmatrix}$ | M1 | Uses their point and their $P$ to find direction of $l_2$ |
| Correct direction for $l_2$ | A1 | |
| $\mathbf{r}=\begin{pmatrix}15\\-1\\25\end{pmatrix}+\mu\begin{pmatrix}7\\-1\\10\end{pmatrix}$ | ddM1 | Attempts vector equation using their direction vector and their point on $l_2$ |
| Correct equation or any equivalent correct vector equation. Must have $\mathbf{r}=$ and not e.g. $l_2=\ldots$ | A1 | |

**Way 3:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal to plane from $l_1$: $\mathbf{r}=\begin{pmatrix}3\\5\\4\end{pmatrix}+t\begin{pmatrix}2\\4\\-1\end{pmatrix}$; $\Rightarrow 2(3+2t)+4(5+4t)-(4-t)=1$; $t=-1\Rightarrow(1,1,5)$ is on $l_2$ | M1 | Complete method to find a point on $l_2$ |
| Direction of $l_2$ is $\begin{pmatrix}15\\-1\\25\end{pmatrix}-\begin{pmatrix}1\\1\\5\end{pmatrix}=\begin{pmatrix}14\\-2\\20\end{pmatrix}$ | M1 | Uses their point and their $P$ to find direction of $l_2$ |
| Correct direction for $l_2$ | A1 | |
| $\mathbf{r}=\begin{pmatrix}1\\1\\5\end{pmatrix}+\mu\begin{pmatrix}7\\-1\\10\end{pmatrix}$ | ddM1 | Attempts vector equation using their direction vector and their point on $l_2$ |
| Correct equation or any equivalent correct vector equation. Must have $\mathbf{r}=$ and not e.g. $l_2=\ldots$ | A1 | |

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