Figure 1 shows a sketch of triangle ABC - Edexcel - A-Level Maths Pure - Question 8 - 2021 - Paper 1

We will use the cosine formula with the dot product:

$\cos ABC = \frac{\vec{AB} \cdot \vec{BC}}{||\vec{AB}|| ||\vec{BC}||}$

First, calculate the dot product ( \vec{AB} \cdot \vec{BC} ):

$\vec{AB} \cdot \vec{BC} = (-3i - 4j - 5k) \cdot (i + j + 4k)$

Calculating this gives:

$= (-3)(1) + (-4)(1) + (-5)(4) = -3 - 4 - 20 = -27$

Next, we find the magnitudes:

$||\vec{AB}|| = \sqrt{(-3)^2 + (-4)^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$

$||\vec{BC}|| = \sqrt{(1)^2 + (1)^2 + (4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$

Now substituting back into the cosine formula:

$\cos ABC = \frac{-27}{(5\sqrt{2})(3\sqrt{2})} = \frac{-27}{30} = -\frac{9}{10}$

As we are interested in the angle, we take the positive value which confirms that:

$\cos ABC = \frac{9}{10}$