What angle does the line pass through the points $A[1,1]$ and $B[−3, −3]$ with the vector $v=(−3,\sqrt{3})$?
I proceed as follows:
I determine the vector $AB = u = (-4, -4)$,
then using the formula
$$\cos (x) = \frac{u\cdot v}{\|u\|\cdot\|v\|}$$
I get
$$\cos (x) = \frac{(-4)\cdot (- 3) + \sqrt{3}\cdot (- 4)}{\sqrt{32} \cdot \sqrt{12}}$$
but somehow I can't make adjustments to the result, which is $\frac{5}{12}\pi$.
Your solution is correct, it just needs some simplification. Your result is
$$\cos(x) = \frac{12-4\sqrt{3}}{\sqrt{32\cdot 12}}$$
while the book result says $x=\frac{5}{12}\pi$. But notice that
$$\cos\left(\frac{5}{12}\pi\right) = \frac{\sqrt{3}-1}{2\sqrt{2}}$$
while $$\frac{12-4\sqrt{3}}{\sqrt{32\cdot 12}} = \frac{4(3-\sqrt{3})}{\sqrt{16}\cdot\sqrt{2\cdot 12}} = \frac{3-\sqrt{3}}{\sqrt{4\cdot 2\cdot 3}} = \frac{\sqrt{3}(\sqrt(3)-1)}{2\cdot\sqrt{2}\cdot\sqrt{3}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$$