Relative to a fixed origin O, the point A has position vector i – 3j + 2k and the point B has position vector –2i + 2j - k - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 6

The vector equation of the line l can be expressed in parametric form. Using point A and the direction vector ( \overrightarrow{AB} ):

Let the position vector ( \overrightarrow{r} ) vary along the line as follows:

$\overrightarrow{r} = \overrightarrow{A} + t \overrightarrow{AB},$

where ( t ) is a scalar parameter. Therefore:

$\overrightarrow{r} = (i - 3j + 2k) + t(-3i + 5j - 3k) \Rightarrow \overrightarrow{r} = (1 - 3t)i + (-3 + 5t)j + (2 - 3t)k.$

Given that AC is perpendicular to l, we can use the scalar product. Let ( \overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} = (2i + pj - 4k) - (i - 3j + 2k) = (1 + p)j - 6k. )

Since ( \overrightarrow{AC} \cdot \overrightarrow{AB} = 0 ), we compute:

$\overrightarrow{AC} \cdot \overrightarrow{AB} = (1 - 3)(1 + p) + 5(0) + (-6)(-3) = 0.$

Expanding this gives:

$-3(1 + p) + 18 = 0 \Rightarrow -3 - 3p + 18 = 0 \Rightarrow 3p = 15 \Rightarrow p = -6.$

To find the distance ( AC ), we first find the position vector for point C:

$\overrightarrow{C} = 2i - 6j - 4k.$

Next, compute ( \overrightarrow{AC} ):

$\overrightarrow{AC} = (2i - 6j - 4k) - (i - 3j + 2k) = (1)i + (-3)j + (-6)k.$

The distance is given by the magnitude of ( \overrightarrow{AC} ):

$|\overrightarrow{AC}| = \sqrt{(1)^2 + (-3)^2 + (-6)^2} = \sqrt{1 + 9 + 36} = \sqrt{46}.$