NO MISREADS ARE ALLOWED IN THIS QUESTION

## Part (a)
$\vec{AB} = \begin{bmatrix}4\\-12\\-20\end{bmatrix}$ | B1 | or $\vec{BA} = \begin{bmatrix}-4\\12\\20\end{bmatrix}$

$\vec{AB}\cdot\begin{bmatrix}3\\-5\\1\end{bmatrix} = (4 \times 3) + (-12 \times -5) + (-20 \times 1)$ | B1ft | Correctly finding scalar product using their $\vec{AB}$ and direction vector of $l_2$ Accept 12 + 60 – 20 or 52

$52 = \sqrt{4^2 + 12^2 + 20^2}\sqrt{3^2 + 5^2 + 1^2}\cos\theta$ | M1 | Correct use of their $\mathbf{a} \cdot \mathbf{b} = a b \cos\theta$

$\cos\theta = \frac{52}{\sqrt{560}\sqrt{35}}$ | A1 | OE all correct in this form or better.

$= \frac{13}{35}$ | A1 | Total: 5 | CAO

The B1 mark for $\vec{AB}$ or $\vec{BA}$ or any multiple could be PI by its use in the scalar product e.g. $\begin{bmatrix}1\\-3\\-5\end{bmatrix}$ etc.

The B1ft mark is for the scalar product of their $\vec{AB}$ with the direction vector of $l_2$.

The M1 mark is for a clear attempt at the scalar product definition of "$\mathbf{a} \cdot \mathbf{b} = a b \cos\theta$" in any form using their $\vec{AB}$ with the direction vector of $l_2$. As in the MS, there is no need for minus signs in squared terms.

Provided each earn the M1 mark with the correct values included, it is possible to score both A1 marks for $\cos\theta = \frac{13}{35}$ without the intermediate form being seen.

## Part (b)
Line AB: $(r =) \begin{bmatrix}0\\6\\9\end{bmatrix} + \mu\begin{bmatrix}4\\-12\\-20\end{bmatrix}$ | M1 | Set up two correct equations for their $l_1$ but with correct $l_2$ and attempt to eliminate $\lambda$ or $\mu$. Accept these three equations in column vector form

$-1 + 3\lambda = 0 + 4\mu$

$5 - 5\lambda = 6 - 12\mu$

$-2 + \lambda = 9 - 20\mu$

$\lambda = 1$ $\mu = \frac{1}{2}$ | A1A1 |

Verifying that all 3 three equations are satisfied and conclusion – e.g. 'intersect'. | E1 | Clear checking of $\lambda$ and $\mu$ in unused equation or showing $P$ lies on both lines. Dependent on **correct co-ordinates of P**.

$P(2,0,-1)$ | A1 | Total: 5 | Accept as a column vector

Look out for any alternative correct versions for the vector equation of $l_1$, e.g. if $B(4, -6, -11)$ is used as known point in $l_1$ this leads to $\lambda = 1$ and $\mu = -\frac{1}{2}$ but also look out for 'multiples' of $\begin{bmatrix}4\\-12\\-20\end{bmatrix}$ being used as the direction vector.

## Part (c)
Using points C and D | |

$\vec{CD} = t\begin{bmatrix}3\\-5\\1\end{bmatrix}$ | B1 | Since C and D lie on $l_2$

$|\vec{AB}| = 3\left|\vec{CD}\right| \Rightarrow \sqrt{560} = 3\sqrt{35t^2}$ | | or $\sqrt{560} = 3t\sqrt{35}$

$t = (\pm)\frac{4}{3}$ | B1 | $\frac{4}{3}$ seen

$\left(\overrightarrow{OC/D}\right) = \vec{OP} + \frac{2}{3}\begin{bmatrix}3\\-5\\1\end{bmatrix}$ | | Uses $P$ the mid-point of $AB$ and $CD$ in this method.

$= \begin{bmatrix}2\\0\\-1\end{bmatrix} + \frac{2}{3}\begin{bmatrix}3\\-5\\1\end{bmatrix}$ or $\begin{bmatrix}2\\0\\-1\end{bmatrix} - \frac{2}{3}\begin{bmatrix}3\\-5\\1\end{bmatrix}$ | M1 | Use of their $\vec{OP}$ and $\frac{2}{3}\begin{bmatrix}3\\-5\\1\end{bmatrix}$

$\left(4, -\frac{10}{3}, -\frac{1}{3}\right)$ | A1 | Accept as a column vector.

$\left(0, \frac{10}{3}, -\frac{5}{3}\right)$ | A1 | Total: 5 | Accept as a column vector.

Candidate could just get $t = \frac{4}{3}$ giving point$\left(4, -\frac{10}{3}, -\frac{1}{3}\right)$ and then find $\left(0, \frac{10}{3}, -\frac{5}{3}\right)$ by symmetry.

## Part (c) continued
$R$ is any point on $l_2$

$(\vec{OR}) = \begin{bmatrix}-1 + 3\lambda\\5 - 5\lambda\\-2 + \lambda\end{bmatrix}$ | |

$(\vec{PR}) = \begin{bmatrix}-1 + 3\lambda\\5 - 5\lambda\\-2 + \lambda\end{bmatrix} - \begin{bmatrix}2\\0\\-1\end{bmatrix}$ | (B1) | $\begin{bmatrix}-3 + 3\lambda\\5 - 5\lambda\\-1 + \lambda\end{bmatrix}$ but award at $\vec{OR} - \vec{OP}$ stage

If $R$ is at C or D then $|\vec{AB}| = 3|\vec{CD}| = 6|\vec{PR}|$ | |

$\sqrt{560} = 6\sqrt{(-3 + 3\lambda)^2 + (5 - 5\lambda)^2 + (-1 + \lambda)^2}$ | (B1) | or better

$560 = 36(35\lambda^2 - 70\lambda + 35)$ | (B1) | |

$9\lambda^2 - 18\lambda + 5 = 0$ | (M1) | Factorising (their $9\lambda^2$ and $+5$ terms) or attempting to solve their quadratic equation correctly – PI by correct values of $\lambda$.

$(3\lambda - 1)(3\lambda - 5) = 0$ | |

$\lambda = \frac{1}{3}$ or $\frac{5}{3}$ | (A1) | Accept as a column vector.

$\left(0, \frac{10}{3}, -\frac{5}{3}\right)$ | (A1) | (5) | Accept as a column vector.

$\left(4, -\frac{10}{3}, -\frac{1}{3}\right)$ | | |

Total | | 15

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