4. The line \(l _ { 1 }\) has vector equation
$$\mathbf { r } = \left( \begin{array} { c }
- 6
4
- 1
\end{array} \right) + \lambda \left( \begin{array} { c }
4
- 1
3
\end{array} \right)$$
and the line \(l _ { 2 }\) has vector equation
$$\mathbf { r } = \left( \begin{array} { c }
- 6
4
- 1
\end{array} \right) + \mu \left( \begin{array} { c }
3
- 4
1
\end{array} \right)$$
where \(\lambda\) and \(\mu\) are parameters.
The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(A\) and the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\) is \(\theta\).
- Write down the coordinates of \(A\).
- Find the value of \(\cos \theta\).
The point \(X\) lies on \(l _ { 1 }\) where \(\lambda = 4\).
- Find the coordinates of \(X\).
- Find the vector \(\overrightarrow { A X }\).
- Hence, or otherwise, show that \(| \overrightarrow { A X } | = 4 \sqrt { } 26\).
The point \(Y\) lies on \(l _ { 2 }\). Given that the vector \(\overrightarrow { Y X }\) is perpendicular to \(l _ { 1 }\),
- find the length of \(A Y\), giving your answer to 3 significant figures.
\section*{LU}