The problem asks us to find the dot product $\vec{a} \cdot \vec{b}$ of two vectors $\vec{a}$ and $\vec{b}$ for two cases: (1) $\vec{a} = (-1, 2)$ and $\vec{b} = (4, 3)$ (2) $\vec{a} = (3, 2)$ and $\vec{b} = (-2, 3)$

1. Problem Description

The problem asks us to find the dot product $\vec{a} \cdot \vec{b}$ of two vectors $\vec{a}$ and $\vec{b}$ for two cases:

(1) $\vec{a} = (-1, 2)$ and $\vec{b} = (4, 3)$

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

2. Solution Steps

The dot product of two vectors $\vec{a} = (a_1, a_2)$ and $\vec{b} = (b_1, b_2)$ is defined as:

$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2$

(1) Given $\vec{a} = (-1, 2)$ and $\vec{b} = (4, 3)$, we have $a_1 = -1$, $a_2 = 2$, $b_1 = 4$, and $b_2 = 3$. Thus,

$\vec{a} \cdot \vec{b} = (-1)(4) + (2)(3) = -4 + 6 = 2$.

(2) Given $\vec{a} = (3, 2)$ and $\vec{b} = (-2, 3)$, we have $a_1 = 3$, $a_2 = 2$, $b_1 = -2$, and $b_2 = 3$. Thus,

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

3. Final Answer

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

(2) $\vec{a} \cdot \vec{b} = 0$