Using the information below: (i) Calculate the angular momentum for the particles. (ii) Find the momentum vectors and then angle between the position.

The angular momentum $\vec{L}$ is the cross product of the position vector $\vec{r}$ and the momentum vector $\vec{p}$.
Given $\vec{r} = (4, 5, 3)$ and $\vec{p} = (1, 4, 2)$.

$$\vec{L} = \vec{r} \times \vec{p} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 5 & 3 \\ 1 & 4 & 2 \end{vmatrix}$$ $$\vec{L} = (5 \cdot 2 - 3 \cdot 4)\mathbf{i} - (4 \cdot 2 - 3 \cdot 1)\mathbf{j} + (4 \cdot 4 - 5 \cdot 1)\mathbf{k}$$ $$\vec{L} = (10 - 12)\mathbf{i} - (8 - 3)\mathbf{j} + (16 - 5)\mathbf{k}$$ $$\vec{L} = -2\mathbf{i} - 5\mathbf{j} + 11\mathbf{k}$$

The angular momentum is $\boxed{(-2, -5, 11)}$.

The dot product is used to find the angle between the vectors.

$$\vec{r} \cdot \vec{p} = (4)(1) + (5)(4) + (3)(2)$$ $$\vec{r} \cdot \vec{p} = 4 + 20 + 6 = 30$$

$$|\vec{r}| = \sqrt{4^2 + 5^2 + 3^2} = \sqrt{16 + 25 + 9} = \sqrt{50}$$ $$|\vec{p}| = \sqrt{1^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$$

The angle $\theta$ is given by $\cos \theta = \frac{\vec{r} \cdot \vec{p}}{|\vec{r}| |\vec{p}|}$.

$$\cos \theta = \frac{30}{\sqrt{50} \sqrt{21}} = \frac{30}{\sqrt{1050}}$$ $$\theta = \arccos\left(\frac{30}{\sqrt{1050}}\right)$$ $$\theta \approx \arccos(0.92582)$$ $$\theta \approx \boxed{22.20^\circ}$$