Relative to a fixed origin O, the point A has position vector $(2i + 3j - 4k)$, the point B has position vector $(4i - 2j + 3k)$, and the point C has position vector $(ai + 5j - 2k)$, where $a$ is a constant and $a < 0$ - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 2

To find the value of a, we need to express the distance |AC| in terms of a.

The vector AC can be expressed as:

$\bar{AC} = \bar{C} - \bar{A} = (ai + 5j - 2k) - (2i + 3j - 4k)$

Calculating this gives:

$\bar{AC} = (a - 2)i + (5 - 3)j + (-2 + 4)k = (a - 2)i + 2j + 2k$

The magnitude of vector AC is given by:

$|\bar{AC}| = \sqrt{(a - 2)^2 + 2^2 + 2^2}$

Substituting the given value:

$4 = \sqrt{(a - 2)^2 + 4 + 4}$

Squaring both sides leads to:

$16 = (a - 2)^2 + 8$

Thus,

$(a - 2)^2 = 8$

Taking the square root gives:

$a - 2 = \pm \sqrt{8} = \pm 2\sqrt{2}$

Solving for a:

1. For $a - 2 = 2\sqrt{2}$,

   $a = 2 + 2\sqrt{2}$ (not valid as $a < 0$)

2. For $a - 2 = -2\sqrt{2}$,

   $a = 2 - 2\sqrt{2}$

Thus, the final answer is:

a = 2 - 2\sqrt{2} ext{ (valid as $a < 0$)}