Vectors u and v are defined by $$u = \begin{pmatrix}-1 \\\ -4 \\\ -3 \end{pmatrix}$$ and $$v = \begin{pmatrix}-7 \\\ 8 \\\ 5 \end{pmatrix}$$ (a) Find u.v - Scottish Highers Maths - Question 2 - 2018

To find the acute angle between two vectors, we can use the formula:

$\cos(\theta) = \frac{u.v}{|u||v|}$

First, we need to calculate the magnitudes of the vectors u and v:

1. The magnitude of u is:

   $|u| = \sqrt{(-1)^2 + (-4)^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$

2. The magnitude of v is:

   $|v| = \sqrt{(-7)^2 + 8^2 + 5^2} = \sqrt{49 + 64 + 25} = \sqrt{138}$

Then we can plug these into the cosine formula:

$\cos(\theta) = \frac{-40}{\sqrt{26} \times \sqrt{138}}$

Calculating the product of the magnitudes:

$\sqrt{26} \times \sqrt{138} = \sqrt{26 \times 138}$

Finally, we solve for \theta:

1. Using the cosine inverse function:

   $\theta = \cos^{-1}\left(\frac{-40}{\sqrt{26 \times 138}}\right)$

2. Determine the value in degrees or radians, noting that we need the acute angle, which will typically be less than 90° or around $\frac{\pi}{2}$ radians.