## Question 5(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2(\lambda-5)+3(-3\lambda-4)-2(5\lambda+3)=6 \Rightarrow \lambda=-2$ then find $x$, $y$, $z$; or e.g. $2x+3(-3x-15-4)-2(5x+25+3)=6 \Rightarrow x=...$ | M1 | Substitutes parametric form of line into plane, solves for parameter, finds at least one coordinate; correct answer only scores no marks |
| $(-7, 2, -7)$ | A1 | Accept as $x=-7,\ y=2,\ z=-7$ or as a vector |

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## Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = \begin{pmatrix}-5\\-4\\3\end{pmatrix} + \begin{pmatrix}2t\\3t\\-2t\end{pmatrix}$ meets the plane when $2(-5+2t)+3(-4+3t)-2(3-2t)=6 \Rightarrow t=...$ | M1 | Identifies a point on $l_1$, uses line through this point perpendicular to plane, finds parameter at intersection; may use different starting point |
| $t=2 \Rightarrow$ mirror point $= \begin{pmatrix}-5\\-4\\3\end{pmatrix}+\begin{pmatrix}2\times4\\3\times4\\-2\times4\end{pmatrix}$ | M1 | Uses twice the parameter to find image point |
| $= \begin{pmatrix}3\\8\\-5\end{pmatrix}$ | A1 | Correct image point |
| $\mathbf{r} = \begin{pmatrix}-7\\2\\-7\end{pmatrix} + \mu\begin{pmatrix}3-(-7)\\8-2\\-5-(-7)\end{pmatrix} = ...$ | ddM1 | Finds equation of line between intersection point from (a) and image point; depends on both previous method marks |
| $\mathbf{r} = \begin{pmatrix}-7\\2\\-7\end{pmatrix} + \mu\begin{pmatrix}10\\6\\2\end{pmatrix}$ | A1* | Fully correct; must have $\mathbf{r}=$; condone different parameter; must be printed form |

**Alternative (first 2 marks):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance from $(-5,-4,3)$ to plane is $\frac{|2\times-5+3\times-4-2\times3-6|}{\sqrt{2^2+3^2+2^2}}=2\sqrt{17}$ | M1 | Identifies point on $l_1$, finds perpendicular distance to plane |
| $\begin{vmatrix}2k\\3k\\-2k\end{vmatrix}=4\sqrt{17} \Rightarrow 4k^2+9k^2+4k^2=16\times17 \Rightarrow k=4$; $k=4 \Rightarrow$ mirror point $= \begin{pmatrix}-5\\-4\\3\end{pmatrix}+\begin{pmatrix}2\times4\\3\times4\\-2\times4\end{pmatrix}$ | M1 | Uses twice distance to find image point |

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## Question 5(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Line joining mirror points intersects plane at $\begin{pmatrix}-5\\-4\\3\end{pmatrix}+\begin{pmatrix}2\times2\\3\times2\\-2\times2\end{pmatrix}$, so equation of line is $\mathbf{r}=\begin{pmatrix}-7\\2\\-7\end{pmatrix}+s\begin{pmatrix}-1-(-7)\\2-2\\-1-(-7)\end{pmatrix}=...$ | M1 | 3.1a |
| $\mathbf{r} = \begin{pmatrix}-7\\2\\-7\end{pmatrix}+s\begin{pmatrix}6\\0\\6\end{pmatrix}$ oe e.g. $\mathbf{r}=\begin{pmatrix}-1\\2\\-1\end{pmatrix}+s\begin{pmatrix}1\\0\\1\end{pmatrix}$ | A1 | |

**Alternative 1 (not on spec):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal to $\Pi_2$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-3&5\\10&6&2\end{vmatrix}=\begin{pmatrix}-36\\48\\36\end{pmatrix}$; Direction of $l_2$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-3&4&3\\2&3&-2\end{vmatrix}=\begin{pmatrix}-17\\0\\-17\end{pmatrix}$ | M1 | |
| $\mathbf{r}=\begin{pmatrix}-7\\2\\-7\end{pmatrix}+s\begin{pmatrix}-17\\0\\-17\end{pmatrix}$ oe | A1 | |

**Alternative 2 (not on spec):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal to $\Pi_2$: $\begin{pmatrix}-36\\48\\36\end{pmatrix}$; $(-3\mathbf{i}+4\mathbf{j}+3\mathbf{k})\cdot(-7\mathbf{i}+2\mathbf{j}-7\mathbf{k})=8$; $\Pi_2$ is $3x-4y-3z=-8$, solve simultaneously with $\Pi_1$, $x=\lambda$ giving $y=2,\ z=\lambda$ | M1 | |
| $\mathbf{r}=\begin{pmatrix}0\\2\\0\end{pmatrix}+s\begin{pmatrix}1\\0\\1\end{pmatrix}$ oe | A1 | |

**Alternative 3 (not on spec):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find equation of plane 2: $3x-4y-3z=-8$; solve simultaneously with plane 1 giving e.g. $y=2,\ x=z$ | M1 | |
| $\mathbf{r}=\begin{pmatrix}s\\2\\s\end{pmatrix}$ oe | A1 | |

## Question 5 (part c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Complete method to find required equation (e.g. finds second point on common line, uses with point from (a) to find direction, forms vector equation) | M1 | May use earlier work or start again |
| Correct equation formed, accept any parameter which is not $x$, $y$ or $z$ | A1 | Must include "$\mathbf{r} =$" unless penalised in (b) |

**Alternative 1:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts normal to $\Pi_2$ via vector product of directions of $l_1$ and $l_2$, then vector product with normal to $\Pi_1$; forms vector equation with point on line | M1 | Normal vectors may also be found using scalar products: $\mathbf{n} = x\mathbf{i}+y\mathbf{j}+\mathbf{k} \Rightarrow (x\mathbf{i}+y\mathbf{j}+\mathbf{k})\cdot(\mathbf{i}-3\mathbf{j}+5\mathbf{k})=0$, $(x\mathbf{i}+y\mathbf{j}+\mathbf{k})\cdot(10\mathbf{i}+6\mathbf{j}+2\mathbf{k})=0$ |
| Correct equation formed, accept any parameter not $x$, $y$ or $z$ | A1 | Must include "$\mathbf{r} =$" unless penalised in (b) |

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## Question 5 (part d):

**Main method:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Direction vector $\begin{pmatrix}1\\0\\1\end{pmatrix}$ perpendicular to $\begin{pmatrix}1\\1\\a\end{pmatrix}$: $\begin{pmatrix}1\\0\\1\end{pmatrix}\cdot\begin{pmatrix}1\\1\\a\end{pmatrix}=0 \Rightarrow 1\times1+0\times1+1\times a=0 \Rightarrow a=\ldots$ | M1 | Realises direction of (c) is perpendicular to normal of $\Pi_3$, applies dot product $= 0$ |
| $a = -1$ | A1 | Correct value for $a$ |
| $b = \begin{pmatrix}-7\\2\\-7\end{pmatrix}\cdot\begin{pmatrix}1\\1\\-1\end{pmatrix}=2$ | A1 | Deduces value of $b$ using their $a$ and one of their points on the line |

**Alternative 1:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-7\\2\\-7\end{pmatrix}\cdot\begin{pmatrix}1\\1\\a\end{pmatrix}=b \Rightarrow -7+2-7a=b$ and $\begin{pmatrix}-1\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}1\\1\\a\end{pmatrix}=b \Rightarrow -1+2-a=b \Rightarrow a=\ldots$ or $b=\ldots$ | M1 | Uses point from (a) and another point on line, substitutes both into given equation |
| $a=-1$ or $b=2$ | A1 | |
| $a=-1$ and $b=2$ | A1 | |

**Alternative 2:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Normal to $\Pi_2$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-3&5\\10&6&2\end{vmatrix}=\begin{pmatrix}-36\\48\\36\end{pmatrix}$; then $\begin{vmatrix}3&-4&-3\\2&3&-2\\1&1&a\end{vmatrix}=0 \Rightarrow 3(3a+2)+4(2a+2)-3(-1)=0 \Rightarrow a=\ldots$ | M1 | Attempts normal to $\Pi_2$, then determinant of matrix of normals $= 0$ |
| $a=-1$ | A1 | |
| $a=-1$ and $b=2$ | A1 | |

**Alternative 3:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{r}=\begin{pmatrix}0\\2\\0\end{pmatrix}+s\begin{pmatrix}1\\0\\1\end{pmatrix}$ lies in $\Pi_3 \Rightarrow s+2+as=b \Rightarrow (a+1)s+2=b \Rightarrow a=\ldots$ | M1 | Substitutes line from (c) into equation for $\Pi_3$, compares coefficients |
| $a=-1$ | A1 | |
| $a=-1$ and $b=2$ | A1 | |

**Alternative 4:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Finds equation for $\Pi_2$: $3x-4y-3z=-8$; solves simultaneously with $\Pi_1$ and $\Pi_3$ using consistency, e.g. $-14a-14=3+3a \Rightarrow a=\ldots$ or $3b+8=42-14b \Rightarrow b=\ldots$ | M1 | Finds $\Pi_2$ equation, solves all 3 equations using consistency |
| $a=-1$ or $b=2$ | A1 | |
| $a=-1$ and $b=2$ | A1 | |

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