# Question 3:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| A: $0 + 6(-2) + 12 = 0$ | B2,1,0 | B1 for two points verified (must see as minimum $-12 + 12 = 0$, $3 - 15 + 12 = 0$, $-12 + 12 = 0$) or any valid complete method for finding or verifying $x + 6y + 12 = 0$ gets M1 A1 |
| B: $3 + 6(-2.5) + 12 = 0$ | | |
| E: $0 + 6(-2) + 12 = 0$ | | |
| At F: $2 + 6a + 12 = 0$ | M1 | Substitution of F into $x + 6y + 12 = 0$ |
| $\Rightarrow 6a = -14$, $a = -14/6 = -7/3$ * | A1 [4] | www NB **AG** |

## Part (ii)(A)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{DH}\cdot\begin{pmatrix}1\\-6\\0\end{pmatrix} = \begin{pmatrix}1\\-6\\0\end{pmatrix}\cdot\begin{pmatrix}0\\0\\3\end{pmatrix} = 1\times0 + (-6)\times0 + 0\times3 = 0$ | B1 | scalar product with a **direction** vector in the plane (including evaluation and $= 0$) (**OR** M1 forms a vector product with at least two correct terms) |
| $\overrightarrow{DC}\cdot\begin{pmatrix}1\\-6\\0\end{pmatrix} = \begin{pmatrix}3\\0.5\\0\end{pmatrix}\cdot\begin{pmatrix}1\\-6\\0\end{pmatrix} = 3\times1 + 0.5\times(-6) + 0\times0 = 0$ | B1 [2] | scalar product with second **direction** vector, with evaluation. (following OR above, A1 correct ie a multiple of $\mathbf{i} - 6\mathbf{j}$) (NB finding only one direction vector and its scalar product is B1 only) |

## Part (ii)(B)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r}\cdot(\mathbf{i} - 6\mathbf{j}) = \mathbf{j}\cdot(\mathbf{i} - 6\mathbf{j})$ | M1 | $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ with $\mathbf{n} = \begin{pmatrix}1\\-6\\0\end{pmatrix}$ and $\mathbf{a} = \begin{pmatrix}0\\1\\3\end{pmatrix}$ or $\begin{pmatrix}0\\1\\0\end{pmatrix}$ or $\begin{pmatrix}0\\1\\0\end{pmatrix}$ or $\begin{pmatrix}3\\1.5\\0\end{pmatrix}$ or substituting H(0,1,3) or D(0,1,0) or C(3,1.5,0) into $x - 6y = d$ |
| $\Rightarrow x - 6y + 6 = 0$ | A1 [2] | oe (isw if $d$ found correctly and $x - 6y = d$ stated). B2 www correct equation stated |

## Part (ii)(C)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2 - 6b + 6 = 0 \Rightarrow b = 4/3$ | B1 | oe – exact answer |
| $FG = 1\frac{1}{3} + 2\frac{1}{3} = 3\frac{2}{3}$ | B1 [2] | oe – exact answer |

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\overrightarrow{FE} =) -2\mathbf{i} + (1/3)\mathbf{j} + \mathbf{k}$, $(\overrightarrow{FB} =) \mathbf{i} - (1/6)\mathbf{j} - 2\mathbf{k}$ | B1 B1 | or $(\overrightarrow{EF} =) 2\mathbf{i} + (-1/3)\mathbf{j} - \mathbf{k}$ or $(\overrightarrow{BF} =) -\mathbf{i} + (1/6)\mathbf{j} + 2\mathbf{k}$ |
| $\cos\theta = \dfrac{-2(1) + (1/3)(-1/6) + 1(-2)}{\sqrt{4 + 1/9 + 1}\sqrt{1 + 1/36 + 4}}$ | M1 | $\cos\theta = (\overrightarrow{FE}\cdot\overrightarrow{FB})/(|\overrightarrow{FE}||\overrightarrow{FB}|)$ (oe) follow through their FE and FB (allow any combination of FE, EF with FB, BF) – allow one sign slip only |
| $\theta = \arccos\left(\dfrac{-73/18}{\frac{\sqrt{46}}{3}\times\frac{\sqrt{181}}{6}}\right)$ | A1 | $\arccos\left(\dfrac{-2 - 1/18 - 2}{5.069}\right) = \arccos(\pm-0.800)$ |
| $\Rightarrow q = 143°$ | A1 [5] | 3sf or better (or 2.5(0) radians or better). Allow candidates who find acute angle using either $\overrightarrow{EF}$ with $\overrightarrow{FB}$ or $\overrightarrow{FE}$ with $\overrightarrow{BF}$ and then state obtuse angle. **Do not isw those who find the obtuse angle and then state the acute angle.** Note: $90 + 2\arctan(1/2)$ is 0/5 |
| **OR** $EF = \sqrt{46}/3$, $FB = \sqrt{181}/6$, $EB = \sqrt{73}/2$ | B3,2,1,0 | One mark for each (2.26, 2.24, 4.27) |
| $\theta = \arccos\left(\dfrac{(\sqrt{46}/3)^2 + (\sqrt{181}/6)^2 - (\sqrt{73}/2)^2}{2(\sqrt{46}/3)(\sqrt{181}/6)}\right)$ | M1 | cosine rule correct with their EF, FB, EB |
| | A1 | $q = 143°$ |

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $z$ coordinate of P is $5/2$ | B1 | stating the correct $z$-coordinate of P; ignore incorrect $x$ and $y$ coordinates (or stated in a position vector) |
| $\overrightarrow{OQ} = \overrightarrow{OP} + \overrightarrow{PQ} = \begin{pmatrix}1\\-13/6\\5/2\end{pmatrix} + \frac{1}{3}\left(\begin{pmatrix}0\\1\\3\end{pmatrix} - \begin{pmatrix}1\\-13/6\\5/2\end{pmatrix}\right)$ | M1 | Complete method for finding the $z$-coordinate of Q or $\overrightarrow{OQ} = \overrightarrow{OH} + (2/3)\overrightarrow{HP}$ or $\overrightarrow{OQ} = (2/3)\overrightarrow{OP} + (1/3)\overrightarrow{OH}$ |
| So height of Q is $8/3$ (metres above ground) | A1 [3] | 2.67 or better |

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