# Question 8:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(6\mathbf{i}+2\mathbf{j}+12\mathbf{k})\cdot(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})=34$ | M1 | Attempt scalar product |
| $\frac{\left|(6\mathbf{i}+2\mathbf{j}+12\mathbf{k})\cdot(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})-5\right|}{\sqrt{3^2+4^2+2^2}}$ | M1 | Use of correct formula |
| $\sqrt{29}$ (not $-\sqrt{29}$) | A1 | Correct distance (allow $29/\sqrt{29}$) |
| | | (3) |

**Way 2:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r}=(6\mathbf{i}+2\mathbf{j}+12\mathbf{k})+\lambda(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})$; $\therefore 6+3\lambda\ \ 3+\ \ 2-4\lambda\ \ -4\ +\ 12+2\lambda\ \ 2=5$ | M1 | Substitutes parametric coordinates of line through $(6,2,12)$ perpendicular to plane into cartesian equation |
| $\lambda=-1 \Rightarrow 3,6,10$ or $-3\mathbf{i}+4\mathbf{j}-2\mathbf{k}$ | M1 | Solves for $\lambda$ to obtain required point or vector |
| $\sqrt{29}$ | A1 | Correct distance |

**Way 3:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Parallel plane containing $(6,2,12)$ is $\mathbf{r}\cdot(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})=34$; $\Rightarrow \frac{\mathbf{r}\cdot(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})}{\sqrt{29}}=\frac{34}{\sqrt{29}}$ | M1 | Origin to this plane is $\frac{34}{\sqrt{29}}$ |
| $\Rightarrow \frac{\mathbf{r}\cdot(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})}{\sqrt{29}}=\frac{5}{\sqrt{29}}$ | M1 | Origin to plane is $\frac{5}{\sqrt{29}}$ |
| $\frac{34}{\sqrt{29}}-\frac{5}{\sqrt{29}}=\sqrt{29}$ | A1 | Correct distance |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&1&5\\1&-1&-2\end{vmatrix}=\begin{pmatrix}3\\9\\-3\end{pmatrix}$ | M1A1 | M1: Attempts $(2\mathbf{i}+\mathbf{j}+5\mathbf{k})\times(\mathbf{i}-\mathbf{j}-2\mathbf{k})$. A1: Any multiple of $\mathbf{i}+3\mathbf{j}-\mathbf{k}$ |
| $\cos\theta=\frac{(3\mathbf{i}-4\mathbf{j}+2\mathbf{k})\cdot(\mathbf{i}+3\mathbf{j}-\mathbf{k})}{\sqrt{3^2+4^2+2^2}\sqrt{1^2+3^2+1^2}}\left(=\frac{-11}{\sqrt{29}\sqrt{11}}\right)$ | M1 | Attempts scalar product of normal vectors including magnitudes |
| $52°$ | dM1 A1 | Obtains angle using arccos (dependent on previous M1). **Do not ISW**: $90°-52°=38°$ loses the A1 |
| | | (5) |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&3&-1\\3&-4&2\end{vmatrix}=\begin{pmatrix}2\\-5\\-13\end{pmatrix}$ | M1A1 | M1: Attempt cross product of normal vectors. A1: Correct vector |
| $x=0:(0,\frac{5}{2},\frac{15}{2})$, $y=0:(1,0,1)$, $z=0:(\frac{15}{13},\frac{-5}{13},0)$ | M1A1 | M1: Valid attempt at a point on both planes. A1: Correct coordinates. May use way 3 to find a point on the line |
| $\mathbf{r}\times(-2\mathbf{i}+5\mathbf{j}+13\mathbf{k})=-5\mathbf{i}-15\mathbf{j}+5\mathbf{k}$ | M1A1 | M1: $\mathbf{r}\times\text{dir}=\text{pos.vector}\times\text{dir}$ (this way round). A1: Correct equation |
| | | (6) |

## Question (c) Way 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| "$x + 3y - z = 0$" and $3x - 4y + 2z = 5$ uses their cartesian form and eliminates $x$, or $y$ or $z$ and substitutes back to obtain two of the variables in terms of the third | M1 | |
| $(x = 1 - \frac{2}{5}y$ and $z = 1 + \frac{13}{5}y)$ or $(y = \frac{5z-5}{13}$ and $x = \frac{15-2z}{13})$ or $(y = \frac{5-5x}{2}$ and $z = \frac{15-13x}{2})$ | A1 | |
| Cartesian Equations: $x = \dfrac{y - \frac{5}{2}}{-\frac{5}{2}} = \dfrac{z - \frac{15}{2}}{-\frac{13}{2}}$ or $\dfrac{x-1}{-\frac{2}{5}} = y = \dfrac{z-1}{\frac{13}{5}}$ or $\dfrac{x - \frac{15}{13}}{-\frac{2}{13}} = \dfrac{y + \frac{5}{13}}{\frac{5}{13}} = z$ | | |
| Points and Directions (direction can be any multiple): $(0, \frac{5}{2}, \frac{15}{2})$, $\mathbf{i} - \frac{5}{2}\mathbf{j} - \frac{13}{2}\mathbf{k}$ or $(1,0,1)$, $-\frac{2}{5}\mathbf{i} + \mathbf{j} + \frac{13}{5}\mathbf{k}$ or $(\frac{15}{13}, -\frac{5}{13}, 0)$, $-\frac{2}{13}\mathbf{i} + \frac{5}{13}\mathbf{j} + \mathbf{k}$ | M1 A1 | M1: Uses their Cartesian equations correctly to obtain a point and direction. A1: Correct point and direction – it may not be clear which is which – look for the correct numbers either as points or vectors |
| Equation of line in required form: $\mathbf{r} \times (-2\mathbf{i} + 5\mathbf{j} + 13\mathbf{k}) = -5\mathbf{i} - 15\mathbf{j} + 5\mathbf{k}$ Or Equivalent | M1 A1 | |
| | **(6)** | **Total 14** |

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## Question (c) Way 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix} 2\lambda + \mu \\ \lambda - \mu \\ 5\lambda - 2\mu \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -4 \\ 2 \end{pmatrix} = 5 \Rightarrow 12\lambda + 3\mu = 5$ | M1A1 | M1: Substitutes parametric form of $\Pi_2$ into the vector equation of $\Pi_1$. A1: Correct equation |
| $\mu = \frac{5}{3}, \lambda = 0$ gives $(\frac{5}{3}, -\frac{5}{3}, \frac{10}{3})$ | M1 | M1: Finds 2 points and direction |
| $\mu = 0, \lambda = \frac{5}{12}$ gives $(\frac{5}{6}, \frac{5}{12}, \frac{25}{12})$ | M1A1 | A1: Correct coordinates and direction |
| Direction $\begin{pmatrix} -2 \\ 5 \\ 13 \end{pmatrix}$ | | |
| Equation of line in required form: $\mathbf{r} \times (-2\mathbf{i} + 5\mathbf{j} + 13\mathbf{k}) = -5\mathbf{i} - 15\mathbf{j} + 5\mathbf{k}$ Or Equivalent | M1A1 | |
| Do not allow 'mixed' methods – mark the best single attempt | | |
| NB for checking, a general point on the line will be of the form: $(1 - 2\lambda,\ 5\lambda,\ 1 + 13\lambda)$ | | |