Given the magnitudes of vectors $\vec{a}$ and $\vec{b}$ as $|\vec{a}| = 1$ and $|\vec{b}| = 3$, and the magnitude of their sum as $|\vec{a} + \vec{b}| = \sqrt{6}$, find the dot product of the two vectors, $\vec{a} \cdot \vec{b}$.

1. Problem Description

Given the magnitudes of vectors $\vec{a}$ and $\vec{b}$ as $|\vec{a}| = 1$ and $|\vec{b}| = 3$, and the magnitude of their sum as $|\vec{a} + \vec{b}| = \sqrt{6}$, find the dot product of the two vectors, $\vec{a} \cdot \vec{b}$.

2. Solution Steps

We know that

$|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$.

Expanding the dot product gives

$|\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b}$.

Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, we have

$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2$.

We are given $|\vec{a}| = 1$, $|\vec{b}| = 3$, and $|\vec{a} + \vec{b}| = \sqrt{6}$. Plugging these values into the equation, we get

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

$6 = 1 + 2(\vec{a} \cdot \vec{b}) + 9$.

$6 = 10 + 2(\vec{a} \cdot \vec{b})$.

$2(\vec{a} \cdot \vec{b}) = 6 - 10$.

$2(\vec{a} \cdot \vec{b}) = -4$.

$\vec{a} \cdot \vec{b} = -2$.

3. Final Answer

$\vec{a} \cdot \vec{b} = -2$