The problem asks to find the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ for three different cases: (1) $\vec{a} = (2, 1)$, $\vec{b} = (1, 3)$ (2) $\vec{a} = (-3, 4)$, $\vec{b} = (4, 3)$ (3) $\vec{a} = (-2, 0)$, $\vec{b} = (3, -\sqrt{3})$

1. Problem Description

The problem asks to find the angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ for three different cases:

(1) $\vec{a} = (2, 1)$, $\vec{b} = (1, 3)$

(2) $\vec{a} = (-3, 4)$, $\vec{b} = (4, 3)$

(3) $\vec{a} = (-2, 0)$, $\vec{b} = (3, -\sqrt{3})$

2. Solution Steps

The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ can be found using the dot product formula:

$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}$$

Thus,

$$\cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$

where $|\vec{a}|$ is the magnitude of vector $\vec{a}$.

(1) $\vec{a} = (2, 1)$, $\vec{b} = (1, 3)$

$\vec{a} \cdot \vec{b} = (2)(1) + (1)(3) = 2 + 3 = 5$

$|\vec{a}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$

$|\vec{b}| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$

$\cos{\theta} = \frac{5}{\sqrt{5} \sqrt{10}} = \frac{5}{\sqrt{50}} = \frac{5}{\sqrt{25 \cdot 2}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$\theta = \arccos{(\frac{\sqrt{2}}{2})} = \frac{\pi}{4}$ or $45^{\circ}$

(2) $\vec{a} = (-3, 4)$, $\vec{b} = (4, 3)$

$\vec{a} \cdot \vec{b} = (-3)(4) + (4)(3) = -12 + 12 = 0$

Since $\vec{a} \cdot \vec{b} = 0$, $\cos{\theta} = 0$.

$\theta = \arccos{(0)} = \frac{\pi}{2}$ or $90^{\circ}$

(3) $\vec{a} = (-2, 0)$, $\vec{b} = (3, -\sqrt{3})$

$\vec{a} \cdot \vec{b} = (-2)(3) + (0)(-\sqrt{3}) = -6 + 0 = -6$

$|\vec{a}| = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2$

$|\vec{b}| = \sqrt{3^2 + (-\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}$

$\cos{\theta} = \frac{-6}{2 \cdot 2\sqrt{3}} = \frac{-6}{4\sqrt{3}} = \frac{-3}{2\sqrt{3}} = \frac{-3\sqrt{3}}{2 \cdot 3} = -\frac{\sqrt{3}}{2}$

$\theta = \arccos{(-\frac{\sqrt{3}}{2})} = \frac{5\pi}{6}$ or $150^{\circ}$

3. Final Answer

(1) $\theta = \frac{\pi}{4}$ or $45^{\circ}$

(2) $\theta = \frac{\pi}{2}$ or $90^{\circ}$

(3) $\theta = \frac{5\pi}{6}$ or $150^{\circ}$