## Question 10(i)(a):

| Point on $l_1$ is $\begin{pmatrix}\lambda a\\9-\lambda\\2+\lambda\end{pmatrix}$ and on $l_2$ is $\begin{pmatrix}-6-\mu a\\-5+2\mu\\10+4\mu\end{pmatrix}$ | B1 | |
|---|---|---|
| Point of intersection: $\begin{pmatrix}\lambda a\\9-\lambda\\2+\lambda\end{pmatrix}=\begin{pmatrix}-6-\mu a\\-5+2\mu\\10+4\mu\end{pmatrix}$ | M1 | Equates coordinates of points |
| $\Rightarrow \mu=1,\ \lambda=12,\ a=-\frac{6}{13}$ | A1 | AG |
| **Total: 3** | | |

## Question 10(i)(b):

Normal to plane using cross product:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\frac{6}{13} & -1 & 1 \\ \frac{6}{13} & 2 & 4 \end{vmatrix} = -6\mathbf{i} + \frac{30}{13}\mathbf{j} - \frac{6}{13}\mathbf{k} \sim -13\mathbf{i} + 5\mathbf{j} - \mathbf{k}$$ | M1 A1 | Uses cross product to find normal to plane AEF |

Using point on plane: e.g. $5(9) - 2 = 43$ | M1 | Substitutes a point |

Equation of plane: $-13x + 5y - z = 43$ | A1 | AEF |

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## Question 10(ii):

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & -1 & 1 \\ -a & 2 & 4 \end{vmatrix} = -6\mathbf{i} - 5a\mathbf{j} + a\mathbf{k}$$

$$\left\| \begin{pmatrix} -6 \\ -5a \\ a \end{pmatrix} \right\| = \sqrt{36 + 26a^2}$$ | B1 | Finds magnitude of cross product of direction vectors of lines |

$$\begin{pmatrix} -6 \\ -5-9 \\ 10-2 \end{pmatrix} \cdot \begin{pmatrix} -6 \\ -5a \\ a \end{pmatrix} = 36 + 78a$$ | M1 A1 | Takes dot product of correct vectors |

$3\sqrt{30}\sqrt{36 + 26a^2} = |36 + 78a|$ | M1 | Puts distance equal to $3\sqrt{30}$ |

$\Rightarrow 15(18 + 13a^2) = (6 + 13a)^2$
$\Rightarrow 26(a-3)^2 = 0 \Rightarrow a = 3$ | A1 | |

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## Question 11E(i):

If $z \neq -1$ then:
$$\frac{z-1}{z+1} = \frac{z^{\frac{1}{2}} - z^{-\frac{1}{2}}}{z^{\frac{1}{2}} + z^{-\frac{1}{2}}} \left(\text{or} = \frac{2i\sin\theta}{2(\cos\theta + 1)}\right)$$ | M1 A1 | Using de Moivre's theorem or multiplying numerator and denominator by $\cos\theta + 1 - i\sin\theta$ |

$$= \frac{2i\sin\frac{1}{2}\theta}{2\cos\frac{1}{2}\theta} = i\tan\frac{1}{2}\theta$$ | A1 | |

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## Question 11E(ii):

If $z = 1$ then $z^r - 1 = 0$ so the given sum is zero. | B1 | |

If $z \neq 1$ then $z = e^{i\frac{2\pi}{3}}$ or $z = e^{-i\frac{2\pi}{3}}$, so:

$$\frac{z^3-1}{z^3+1} + \frac{z^2-1}{z^2+1} + \frac{z-1}{z+1} = 0 + \frac{e^{\pm i\frac{4\pi}{3}}-1}{e^{\pm i\frac{4\pi}{3}}+1} + \frac{e^{\pm i\frac{2\pi}{3}}-1}{e^{\pm i\frac{2\pi}{3}}+1}$$ | B1 M1 A1 | Must consider all three roots of unity; AEF two of three terms correct for M1 |

$$\frac{e^{\pm i\frac{4\pi}{3}}-1}{e^{\pm i\frac{4\pi}{3}}+1} + \frac{e^{\pm i\frac{2\pi}{3}}-1}{e^{\pm i\frac{2\pi}{3}}+1} = i\tan\frac{1}{2}\left(\frac{4\pi}{3}\right) + i\tan\frac{1}{2}\left(\frac{2\pi}{3}\right) = 0$$ | A1 | AG |

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## Question 11E(iii):

$z = 1,\ e^{\pm i\frac{2\pi}{3}}$,
$(z^3-1)(z^2+1)(z+1) + (z^2-1)(z^3+1)(z+1) + (z-1)(z^3+1)(z^2+1) = 0$ | B1 | Writes cube roots of unity, AEF, must be exact |

$\Rightarrow 3z^6 + z^5 + z^4 - z^2 - z - 3 = 0$ | M1 | Expands |

$3z^6 + z^5 + z^4 - z^2 - z - 3 = (z+1)(z^3-1)(3z^2-2z+3)$ | M1 A1 | Factorises |

$\Rightarrow z = -1,\quad \frac{2 \pm i\sqrt{32}}{6} = \frac{1}{3} \pm i\frac{2}{3}\sqrt{2}$ | M1 A1 | Finds other three roots, AEF, must be exact |

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## Question 11O(i):

e.g. $2\mathbf{e}$ | B1 | Allow any scalar multiple $\mu\mathbf{e}$ where $\mu \neq 0, 1$ |

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## Question 11O(ii):

Eigenvector: $\mathbf{e}$, Eigenvalue: $\lambda^n$ | B1 B1 | Allow any scalar multiple $\mu\mathbf{e}$ where $\mu \neq 0$; Note: $A^n\mathbf{e} = \lambda^n\mathbf{e}$ SCB1 |

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## Question 11O(iii):

Eigenvalues of diagonal matrix $\mathbf{A}$: $\lambda = 3, 7, 1$
(Or from characteristic equation: $(\lambda-3)(\lambda-7)(\lambda-1)=0$) | B1 | |

$\lambda = 3$: $\mathbf{e}_1 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&4&0\\4&8&-2\end{vmatrix} = \begin{pmatrix}-8\\4\\0\end{pmatrix} = t\begin{pmatrix}-2\\1\\0\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |

$\lambda = 7$: $\mathbf{e}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&0&0\\4&8&-6\end{vmatrix} = \begin{pmatrix}0\\-24\\-32\end{pmatrix} = t\begin{pmatrix}0\\3\\4\end{pmatrix}$ | A1 | |

$\lambda = 1$: $\mathbf{e}_3 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&6&0\\4&8&0\end{vmatrix} = \begin{pmatrix}0\\0\\-8\end{pmatrix} = t\begin{pmatrix}0\\0\\1\end{pmatrix}$ | A1 | |

$$\mathbf{P} = \begin{pmatrix}-2&0&0\\1&3&0\\0&4&1\end{pmatrix} \text{ and } \mathbf{D} = \begin{pmatrix}3^n&0&0\\0&7^n&0\\0&0&1\end{pmatrix}$$ | B1 FT B1 FT | Or correctly matched permutations of columns; FT on non-zero and distinct eigenvalues and vectors |

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## Question 11O(iv):

$$\sum_{n=1}^{N}(k^n A^n - k^{n+1}A^{n+1}) = kA - k^{N+1}A^{N+1}$$ | M1 A1 | Method of differences |

$k^{N+1}A^{N+1} \to 0$ as $N \to \infty$ for | M1 | |

$-\frac{1}{7} < k < \frac{1}{7}$ | A1 | |