With respect to a fixed origin O, the lines l_1 and l_2 are given by the equations l_1 : r = (4, 28, 4) + \ (-1, -5, 1) \lambda\n l_2 : r = (5, 3, 1) + \ (0, 3, -4) \mu where \lambda and \mu are scalar parameters - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 5

To find the angle \theta between the lines l_1 and l_2, we can use the dot product formula:

$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| ||\mathbf{b}||}$

where \mathbf{a} = (-1, -5, 1) and \mathbf{b} = (0, 3, -4).

Calculating the dot product: $\mathbf{a} \cdot \mathbf{b} = (-1)(0) + (-5)(3) + (1)(-4) = -15 - 4 = -19$

Calculating the magnitudes: $||\mathbf{a}|| = \sqrt{(-1)^2 + (-5)^2 + (1)^2} = \sqrt{27}$ $||\mathbf{b}|| = \sqrt{(0)^2 + (3)^2 + (-4)^2} = \sqrt{25} = 5$

Now plugging these into the formula: $\cos \theta = \frac{-19}{\sqrt{27} \cdot 5}$

Calculating \theta yields: $\theta = \cos^{-1}\left(-\frac{19}{5 \sqrt{27}}\right)\approx 120.63 \text{ degrees}$

The acute angle is thus 120.63 degrees.

The position vector of point A is \begin{pmatrix} 2 \ 18 \ 6 \end{pmatrix} and the coordinates of X are (5, 3, 1).

The distance AX can be calculated using: $AX = ||\overrightarrow{AX}|| = ||\begin{pmatrix} 5 - 2 \\ 3 - 18 \\ 1 - 6 \end{pmatrix}|| = ||\begin{pmatrix} 3 \\ -15 \\ -5 \end{pmatrix}||$

Calculating this magnitude: $AX = \sqrt{3^2 + (-15)^2 + (-5)^2} = \sqrt{9 + 225 + 25} = \sqrt{259}$

So the distance AX is given as \sqrt{259}.

Since \overrightarrow{YA} is perpendicular to line l_1, we can find point Y as follows.

Let point Y be represented by \begin{pmatrix} 5 + 0\mu \ 3 + 3\mu \ 1 - 4\mu \end{pmatrix}.

Using the vector conditions: $\overrightarrow{YA} \cdot \mathbf{a} = 0$

Where \overrightarrow{YA} = \begin{pmatrix} 2 - (5 + 0\mu) \ 18 - (3 + 3\mu) \ 6 - (1 - 4\mu) \end{pmatrix}.

Solving \overrightarrow{YA} \cdot \mathbf{a} = 0 leads to: $\mathbf{a} = (-1,-5,1)$ $\begin{pmatrix} -3 - 0\mu \\ 15 - 3\mu \\ 5 + 4\mu \end{pmatrix} \cdot \begin{pmatrix} -1 \\ -5 \\ 1 \end{pmatrix} = 0$

Solving this yields the necessary value for \mu, and substituting that back into the position gives Y.

Using the distance formula: $YA = ||\overrightarrow{YA}||$ yields the numerical value rounded to one decimal place.

Given that |\overrightarrow{AX}| = 2|\overrightarrow{AB}|,

Assuming \overrightarrow{AB} = \begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix},

we have: \overrightarrow{AX} = \begin{pmatrix} 3 \ -15 \ -5 \end{pmatrix} \ = 2 \begin{pmatrix} b_1 \ b_2 \ b_3 \end{pmatrix} .

The above forms the basis for two linear equations to derive B's two position vectors. Substituting the vector values provides:

1. |\overrightarrow{B}| = 2 \text{ for the first position vector.}
2. |\overrightarrow{B}| = -2 \text{ for the second position vector.}

The values of B hence can be obtained from inputting the derived values.