**(i)** $(\mathbf{i} - (2\sin t)\mathbf{j}) \times (4\mathbf{j} - \mathbf{k}) = (2\sin t)\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ | M1A1 |

$PQ = (\mathbf{i} - \mathbf{j}). [(2\sin t)\mathbf{i} + \mathbf{j} + 4\mathbf{k}]/\sqrt{4\sin^2 t + 17}$ | dM1A1 |

$PQ = |1 - 2\sin t|/\sqrt{4\sin^2 t + 17}$ | A1 |

Condone disappearing modulus sign –

Deduct 1 mark if no modulus sign at all |

**(ii)** $PQ = 0 \Rightarrow \sin t = 1/2 \Rightarrow t = \pi/6, 5\pi/6$ or $0.524, 2.62$ | M1A1 (both) |

**(iii)** Obtains some vector $\perp$ plane $BPQ$, e.g., | M1A1 |

$(\sqrt{2}\mathbf{i} + \mathbf{j} + 4\mathbf{k}) \times (\mathbf{i} - \sqrt{2}\mathbf{j}) = 4\sqrt{2}\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$ |

$\mathbf{p}(A: BPQ) = (4\sqrt{2}\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}) \cdot (\mathbf{i} - \mathbf{j})/\sqrt{57}$ | M1A1 |

$= 4(\sqrt{2} - 1)/\sqrt{57} = 0.219$ | A1 |

**or**

**For (i)**

$\ell_1 = r\begin{pmatrix}1\\4\end{pmatrix} + \lambda\begin{pmatrix}0\\-1\end{pmatrix}$ $\Rightarrow \overline{PQ} = \begin{pmatrix}1 - 4\lambda - 2\mu\sin t\\\end{pmatrix}$ |

$\ell_2 = r\begin{pmatrix}1\\2\\4\end{pmatrix} + \mu\begin{pmatrix}-2\sin t\\0\end{pmatrix}$ |

$\overline{PQ} \cdot \begin{pmatrix}0\\-1\end{pmatrix} = 0$ $\Rightarrow 17\lambda + 8\mu\sin t = 4$ |

$\overline{PQ} \cdot \begin{pmatrix}1\\-2\sin t\\0\end{pmatrix} = 0$ $\Rightarrow 8\sin t\lambda + (1 + 4\sin^2 t)\mu = 1 + 2\sin t$ | M1 |

Whence $\lambda = \frac{4(1 - 2\sin t)}{(17 + 4\sin^2 t)}$ and $\mu = \frac{17 + 2\sin t}{(17 + 4\sin^2 t)}$ | A1 (both) |

For $\overline{PQ}$ $x = \frac{2\sin t(1 - 2\sin t)}{(17 + 4\sin^2 t)}$, $y = \frac{1 - 2\sin t}{(17 + 4\sin^2 t)}$, $z = \frac{4(1 - 2\sin t)}{17 + 4\sin^2 t}$ | dM1A1 |

($\lambda$ and $\mu$ of the form $\frac{a + b\sin t}{17 + 4\sin^2 t}$) |

$|PQ| = \sqrt{x^2 + y^2 + z^2} = \frac{|1 - 2\sin t|}{17 + 4\sin^2 t}}\sqrt{4\sin^2 t + 1^2 + 4^2}$ | A1 |

$= \frac{|1 - 2\sin t|}{\sqrt{17 + 4\sin^2 t}}$ |

**For (iii) Eliminates $\lambda$ and $\mu$ from parametric equation of plane $BPQ$**

$\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2\\4\end{pmatrix} + \lambda\begin{pmatrix}-\sqrt{2}\\-\sqrt{2}\end{pmatrix} + \mu\begin{pmatrix}\sqrt{2}\\1\\4\end{pmatrix}$ | M1A1 |

to obtain cartesian equation of plane $BPQ$ |

$4\sqrt{2}x + 4y - 3z = 4(\sqrt{2} - 1)$ * | A1 |

Uses distance of point from a line formula with * and $(2, 1, 4)$ | M1 |

$d = \frac{8\sqrt{2} + 4 - 12 - 4\sqrt{2} + 4}{{\sqrt{32 + 16 + 9}}} = \frac{4(\sqrt{2} - 1)}{\sqrt{57}}$ |

$= 0.219$ cao | A1 |

# Question 12 - EITHER

$(1 + i\tan\theta)^n = \sec^n\theta(\cos k\theta + i\sin k\theta)$ | M1A1 |

$\sum_{k=0}^{n-1}(1 + i\tan\theta)^k = [(1 + i\tan\theta)^n - 1]/i\tan\theta$ | M1A1 |

$= \sec^n\theta\sin n\theta\cot\theta + i(\cot\theta - \sec^n\theta\cos n\theta\cot\theta)$ | A1 |

$\Rightarrow \sum_{k=0}^{n-1}\sec^k\theta\cos k\theta = \sec^n\theta\sin n\theta\cot\theta$ (AG) | M1A1 |

Put $\theta = \pi/3$ to obtain second result (AG) | M1A1 |

$x = \cos\theta \Rightarrow \sec\theta = 1/x$, $\cot\theta = x/\sqrt{1 - x^2}$ | M1A1 |

Uses these results to obtain final result (AG) | M1A1 |

**First 7 marks**

$\sum_0^{n-1}(1 + i\tan\theta)^k = \sum_0^{n-1}(\cos k\theta + i\sin k\theta)\sec^k\theta = \frac{\{1 - (1 + i\tan\theta)^n\}}{-i\tan\theta}$ | M1A1 | M1A1 |

$\therefore\sum_0^{n-1}\cos k\theta\sec^k\theta = \text{Re}\left\{\frac{1 - (\cos n\theta + i\sin n\theta)\sec^n\theta}{i\tan\theta}\right\}$ | M1A1 |

$= \sin n\theta\sec^n\theta\cot\theta$ | (AG) | A1 |

# Question 12 - OR

**(i)** Reduces $M_1$ to col echelon form by elementary col operations

or $M_1^T$ to row echelon form by elementary row operations | M1 |

$M_1^T \to \begin{pmatrix}3 & 0 & 0\\0 & 3 & 0\\0 & 0 & 1\\1 & 7 & -1\end{pmatrix}$ hence $\begin{pmatrix}3\\0\\0\\1\end{pmatrix}, \begin{pmatrix}0\\3\\0\\7\end{pmatrix}, \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$ | M1 |

Any correct echelon form | A1 |

Selects 3 li cols from reduced $M_1$ (or equiv from reduced $M_1^T$) | M1 |

(Allow M1A1M1 for any other valid method) |

Any correct basis of $R_1$, e.g., $\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \begin{pmatrix}0\\3\\6\\1\end{pmatrix}, \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$ or $\begin{pmatrix}3\\0\\0\\-1\end{pmatrix}, \begin{pmatrix}0\\3\\0\\7\end{pmatrix}, \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$ | A1 |

**or for (i)**

Reduces $M_1$ to row echelon form by elementary row operations | M1 |

$\begin{pmatrix}1 & 1 & 1 & 2\\0 & 1 & 2 & 2\\0 & 0 & 2 & 1\\0 & 0 & 0 & 0\end{pmatrix}$ | M1 |

Any correct row echelon form | A1 |

No working for echelon matrix here, or in (ii) gets B1 |

$rr(M_1) = 3 \Rightarrow$ any 3 li columns form a basis of $M_1$ | M1 |

May be implied by answer |

Any correct basis of $R_1$, e.g., $\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \begin{pmatrix}1\\4\\7\\2\end{pmatrix}, \begin{pmatrix}1\\7\\11\\5\end{pmatrix}$ | A1 |

**or**

$\det M_1 = 0$ (calculator) | M1 |

$\Rightarrow rM_1 \leq 3$ | A1 |

First 3 columns of $M_1$ are clearly linearly independent | M1 |

Hence basis | A1 |

**(ii)** Reduces $M_2$ to echelon form by elementary row operations | M1 |

$\begin{pmatrix}2 & 0 & -1 & -1\\0 & 2 & -1 & -1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{pmatrix}$ | M1 |

Any correct row echelon form | A1 |

Valid method to find basis of $K_2$ | M1 |

(Allow M1A1M1 for any other valid method) |

Any correct basis of $K_2$, e.g., $\begin{pmatrix}1\\1\\0\\2\end{pmatrix}, \begin{pmatrix}1\\1\\2\\0\end{pmatrix}$ or $\begin{pmatrix}1\\0\\1\\1\end{pmatrix}, \begin{pmatrix}1\\0\\1\\-1\end{pmatrix}$ | A1 |

Shows each basis element of $K_2$ is in $R_1$ (AG) | B1 |

**First 4 marks**

$M_2^T = \begin{pmatrix}2 & 5 & 3 & 13\\a\\0 & 1 & -1 & -1\\b\\-1 & -3 & -1 & -6\\c\\-1 & -3 & -1 & -6\end{pmatrix}d \to \begin{pmatrix}2 & 5 & 3 & 13\\a\\0 & 1 & -1 & -1\\b\\0 & 0 & 0 & a+b+2c\\0 & 0 & 0 & c-d\end{pmatrix}$ | M1A1 |

Basis is $\begin{pmatrix}1\\1\\2\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$ | M1A1 |

**(iii)** Any valid argument, e.g., $W$ does not contain zero vector so $W$ not a vector space | B1 |

**(iv)** For any vector **x**, $\mathbf{M_2M_1}\mathbf{x} = \mathbf{M_2}(\alpha\mathbf{b_1} + \beta\mathbf{b_2} + \gamma\mathbf{b_3})$, where $\mathbf{b_1}, \mathbf{b_2}, \mathbf{b_3}$ are any 3 l.i. basis vectors of $R_1$, 2 of which must be basis vectors of $K_2$ | M1 |

Hence if $\mathbf{b_1}$ and $\mathbf{b_2}$ are basis vectors of $K_2$, then $\mathbf{M_2M_1}\mathbf{x} = \gamma\mathbf{M_2}\mathbf{b_3}$ | A1 |

Hence as $\dim(\text{range of }T_3) = 1$, then the dimension of the null space of $T_3 = 4 - 1 = 3$ | A1 |

$\text{or } \mathbf{M_2M_1} = \begin{pmatrix}0 & -7 & -14 & -14\\0 & -18 & -36 & -36\\0 & -10 & -20 & -20\\0 & -45 & -90 & -90\end{pmatrix}$ | B1 |

Nullity $= 4 - r(\mathbf{M_2M_1}) = 3$ | M1A1 |