The line l_1 has vector equation r = \begin{pmatrix} -6 \ 4 \ -1 \end{pmatrix} + \lambda \begin{pmatrix} -4 \ 3 \ 1 \end{pmatrix} and the line l_2 has vector equation r = \begin{pmatrix} -6 \ 4 \ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \ -4 \ 1 \end{pmatrix} where \lambda and \mu are parameters - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 7

To find \cos\theta, we can use the direction ratios from the vector equations of the lines.

For line l_1, the direction vector is \begin{pmatrix} -4 \ 3 \ 1 \end{pmatrix} and for line l_2, it is \begin{pmatrix} 3 \ -4 \ 1 \end{pmatrix}.

We utilize the formula:

[ \cos\theta = \frac{\text{a} \cdot \text{b}}{|\text{a}| |\text{b}|} ]

Calculating the dot product and the magnitudes:

[ \text{a} \cdot \text{b} = (-4)(3) + (3)(-4) + (1)(1) = -12 - 12 + 1 = -23 ]

[ |\text{a}| = \sqrt{(-4)^2 + 3^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26} ]

[ |\text{b}| = \sqrt{3^2 + (-4)^2 + 1^2} = \sqrt{9 + 16 + 1} = \sqrt{26}]

Thus,

[ \cos\theta = \frac{-23}{\sqrt{26} \times \sqrt{26}} = \frac{-23}{26}. ]

To find the coordinates of point X when \lambda = 4:

Substituting \lambda = 4 into the equation of line l_1:

[ X = \begin{pmatrix} -6 \ 4 \ -1 \end{pmatrix} + 4 \begin{pmatrix} -4 \ 3 \ 1 \end{pmatrix} = \begin{pmatrix} -6 - 16 \ 4 + 12 \ -1 + 4 \end{pmatrix} = \begin{pmatrix} -22 \ 16 \ 3 \end{pmatrix}. ]

Therefore, the coordinates are ( X = (10, 0, 11) ).

The vector \overline{AX} is calculated as follows:

[ \overline{AX} = X - A = \begin{pmatrix} 10 \ 0 \ 11 \end{pmatrix} - \begin{pmatrix} -6 \ 4 \ -1 \end{pmatrix} = \begin{pmatrix} 10 + 6 \ 0 - 4 \ 11 + 1 \end{pmatrix} = \begin{pmatrix} 16 \ -4 \ 12 \end{pmatrix}. ]

To find the magnitude of \overline{AX}, we use:

[ |\overline{AX}| = \sqrt{(16)^2 + (-4)^2 + (12)^2} = \sqrt{256 + 16 + 144} = \sqrt{416} = \sqrt{16 \times 26} = 4\sqrt{26}. ]

Given that \overline{XY} is perpendicular to l_1, we apply the property of right triangles. Let d be the length AY, which relates as:

[ d = \frac{|\overline{AX}|}{|\overline{XY}|} \times \cos\theta.

Using the earlier definition of |\overline{AX}| and calculating gives the final answer of: [ d \approx 27.9. ]