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 given by the cross product of the position vector $\vec{r}$ and the linear momentum vector $\vec{p}$:

$$\vec{L} = \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 1 & 4 & -2 \end{vmatrix}$$ $$\vec{L} = (5(-2) - 3(4))\hat{i} - (4(-2) - 3(1))\hat{j} + (4(4) - 5(1))\hat{k}$$ $$\vec{L} = (-10 - 12)\hat{i} - (-8 - 3)\hat{j} + (16 - 5)\hat{k}$$ $$\vec{L} = -22\hat{i} + 11\hat{j} + 11\hat{k}$$

The angular momentum is $\boxed{\vec{L} = (-22, 11, 11)}$.

The momentum vector is given as $\vec{p} = (1,4,-2)$.
To find the angle $\theta$ between $\vec{r}$ and $\vec{p}$, we use the dot product formula:

$$\cos \theta = \frac{\vec{r} \cdot \vec{p}}{|\vec{r}| |\vec{p}|}$$

First, calculate the dot product:

$$\vec{r} \cdot \vec{p} = (4)(1) + (5)(4) + (3)(-2) = 4 + 20 - 6 = 18$$

Next, calculate the magnitudes of $\vec{r}$ and $\vec{p}$:

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

Now, substitute these values into the cosine formula:

$$\cos \theta = \frac{18}{\sqrt{50} \sqrt{21}} = \frac{18}{\sqrt{1050}}$$ $$\theta = \arccos\left(\frac{18}{\sqrt{1050}}\right) \approx \arccos(0.55549) \approx 56.26^\circ$$

The angle between the position and momentum vectors is $\boxed{\theta \approx 56.26^\circ}$.

$$\vec{L} = \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 7 & -1 & 1 \end{vmatrix}$$ $$\vec{L} = ((-2)(1) - 3(-1))\hat{i} - (1(1) - 3(7))\hat{j} + (1(-1) - (-2)(7))\hat{k}$$ $$\vec{L} = (-2 + 3)\hat{i} - (1 - 21)\hat{j} + (-1 + 14)\hat{k}$$ $$\vec{L} = 1\hat{i} + 20\hat{j} + 13\hat{k}$$

The angular momentum is $\boxed{\vec{L} = (1, 20, 13)}$.

The momentum vector is given as $\vec{p} = (7,-1,1)$.
First, calculate the dot product:

$$\vec{r} \cdot \vec{p} = (1)(7) + (-2)(-1) + (3)(1) = 7 + 2 + 3 = 12$$

Next, calculate the magnitudes of $\vec{r}$ and $\vec{p}$:

$$|\vec{r}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$ $$|\vec{p}| = \sqrt{7^2 + (-1)^2 + 1^2} = \sqrt{49 + 1 + 1} = \sqrt{51}$$

Now, substitute these values into the cosine formula:

$$\cos \theta = \frac{12}{\sqrt{14} \sqrt{51}} = \frac{12}{\sqrt{714}}$$ $$\theta = \arccos\left(\frac{12}{\sqrt{714}}\right) \approx \arccos(0.44983) \approx 63.27^\circ$$

The angle between the position and momentum vectors is $\boxed{\theta \approx 63.27^\circ}$.

$$\vec{L} = \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 0 \\ 1 & 0 & 0 \end{vmatrix}$$ $$\vec{L} = (2(0) - 0(0))\hat{i} - (0(0) - 0(1))\hat{j} + (0(0) - 2(1))\hat{k}$$ $$\vec{L} = (0 - 0)\hat{i} - (0 - 0)\hat{j} + (0 - 2)\hat{k}$$ $$\vec{L} = 0\hat{i} + 0\hat{j} - 2\hat{k}$$

The angular momentum is $\boxed{\vec{L} = (0, 0, -2)}$.

The momentum vector is given as $\vec{p} = (1,0,0)$.
First, calculate the dot product:

$$\vec{r} \cdot \vec{p} = (0)(1) + (2)(0) + (0)(0) = 0 + 0 + 0 = 0$$

Since the dot product is 0, the vectors $\vec{r}$ and $\vec{p}$ are orthogonal.

$$\cos \theta = \frac{0}{|\vec{r}| |\vec{p}|} = 0$$ $$\theta = \arccos(0) = 90^\circ$$

The angle between the position and momentum vectors is $\boxed{\theta = 90^\circ}$.