The line $l_1$ has equation $$egin{pmatrix} 1 \ 0 \ -1 \\ 0 \ 1 \ 0 \\ 0 \ 0 \ 1 \\ \\ 0 \ -1 \ 0 \ \end{pmatrix} + \lambda \begin{pmatrix} 1 \ 1 \ 0 \ \end{pmatrix}.$$ The line $l_2$ has equation $$egin{pmatrix} 1 \ 3 \ 6 \\ 2 \ 6 \ -1 \\ -1 \ -1 \ 1 \\ \\ 1 \ 0 \ 1 \ \end{pmatrix} + \mu \begin{pmatrix} 1 \ 1 \ -1 \\ \end{pmatrix}.$$ (a) Show that $l_1$ and $l_2$ do not meet - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 7

First, we find the coordinates of points $A$ and $B$. Given $\lambda = 1$, the coordinates of point $A$ on line $l_1$ are:
$A = \begin{pmatrix} 1 + 1 \ \ 0 + 1 \ \ -1 \ \end{pmatrix} = \begin{pmatrix} 2 \ 1 \ -1 \ \end{pmatrix}.$
For $\mu = 2$, the coordinates of point $B$ on line $l_2$ are:
$B = \begin{pmatrix} 1 + 2 \ \ 3 + 2 \ \ 6 \ \end{pmatrix} = \begin{pmatrix} 3 \ 5 \ 6 \ \end{pmatrix}.$
Next, we find vector $\overrightarrow{AB}$:
$\overrightarrow{AB} = B - A = \begin{pmatrix} 3 \ 5 \ 6 \ \end{pmatrix} - \begin{pmatrix} 2 \ 1 \ -1 \ \end{pmatrix} = \begin{pmatrix} 1 \ 4 \ 7 \ \end{pmatrix}.$
The direction vector of line $l_1$ is $\begin{pmatrix} 1 \ 1 \ 0 \ \end{pmatrix}$.
Now we calculate the cosine of the angle $\theta$ between these vectors using the dot product:
$\cos(\theta) = \frac{\overrightarrow{AB} \cdot d_{l_1}}{|\overrightarrow{AB}| |d_{l_1}|}.$
Calculating the dot product:
$\overrightarrow{AB} \cdot d_{l_1} = \begin{pmatrix} 1 \ 4 \ 7 \ \end{pmatrix} \cdot \begin{pmatrix} 1 \ 1 \ 0 \ \end{pmatrix} = 1 \cdot 1 + 4 \cdot 1 + 7 \cdot 0 = 5.$
Calculating the magnitudes:
$|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + 7^2} = \sqrt{66}, \quad |d_{l_1}| = \sqrt{1^2 + 1^2} = \sqrt{2}.$
Putting it all together:
$\cos(\theta) = \frac{5}{\sqrt{66} \cdot \sqrt{2}} = \frac{5}{\sqrt{132}}.$ Finally, the cosine of the acute angle between $AB$ and $l_1$ is given by:
$$\cos(\theta) \approx 0.435.$