With respect to a fixed origin O, the line l_1 is given by the equation $$ r = \begin{pmatrix} 8 \\ 1 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} $$ where \( \mu \) is a scalar parameter - Edexcel - A-Level Maths Pure - Question 2 - 2016 - Paper 4

The line l_2 passes through point P and is parallel to line l_1. Therefore, the vector equation for line l_2 can be written as:

$l_2: \quad r = \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix}$ where ( \lambda ) is a scalar parameter.

The distance AP can be calculated using the distance formula:

$AP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substituting the coordinates of points A and P:

$AP = \sqrt{(3 - 1)^2 + (5 - 5)^2 + (0 - 2)^2} = \sqrt{(2)^2 + (0)^2 + (-2)^2} = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2}.$

Thus, the value of k is 2.

To find ( \cos \theta ), where ( \theta ) is the angle between vectors AP and l_1, we use the dot product:

$\cos \theta = \frac{AP \cdot D}{|AP| |D|}$

Where D is the direction vector of line l_1, ( D = \begin{pmatrix} -5 \ 4 \ 3 \end{pmatrix} ).

Thus:

$AP = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix}$

Calculating the dot product:

$AP \cdot D = (2)(-5) + (0)(4) + (-2)(3) = -10 - 6 = -16.$

The magnitudes are:

$|AP| = 2\sqrt{2}, \quad |D| = \sqrt{(-5)^2 + (4)^2 + (3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}.$

Calculating ( \cos \theta ):

$\cos \theta = \frac{-16}{(2\sqrt{2})(5\sqrt{2})} = \frac{-16}{20} = -0.8.$

The area of triangle APE can be found using:

$Area = \frac{1}{2} \times base \times height$

Using the distance AP as the base and height based on the sine of the angle between AP and l_2:

Thus,

$Area = \frac{1}{2} \times 2\sqrt{2} \times |AP| \times sin \theta = \frac{1}{2} \times 2\sqrt{2} \times 2.4 = 2.4.$

The coordinates of point E can be found knowing that the distance AP equals PE. Therefore, we can set:

$PE = \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} \lambda$

Thus,

Position 1: ( E = A + \begin{pmatrix} -5 \ 4 \ 3 \end{pmatrix} \lambda_1 )

Position 2: ( E = A - \begin{pmatrix} -5 \ 4 \ 3 \end{pmatrix} \lambda_2 )

Calculating these will yield the two positions for E.