I am given the following problem:
Let $\Vert \overrightarrow{a}\Vert = 3$ , $\Vert \overrightarrow{b}\Vert = 2$ and $\angle \left(\overrightarrow{a},\overrightarrow{b}\right) = 60^\circ$. Find $\Vert \overrightarrow{a}+\overrightarrow{b}\Vert$, $\Vert \overrightarrow{a}-\overrightarrow{b}\Vert$, $\overrightarrow{a} \cdot \overrightarrow{b}$ and $\angle \left(\overrightarrow{a}+\overrightarrow{b}, \overrightarrow{a}-\overrightarrow{b}\right)$.
I was able to evaluate $\Vert \overrightarrow{a}+\overrightarrow{b}\Vert$ and $\Vert \overrightarrow{a}-\overrightarrow{b}\Vert$ through cosine Law:
$$ \Vert \overrightarrow{a}+\overrightarrow{b}\Vert = \sqrt{19}\\ \Vert \overrightarrow{a}-\overrightarrow{b}\Vert = \sqrt{7} $$
I am stuck on the next step. If I could evaluate the angle between $\Vert \overrightarrow{a}+\overrightarrow{b}\Vert$ and $\Vert \overrightarrow{a}-\overrightarrow{b}\Vert$ I could finish the exercise. If somebody could give me a hint, I'll appreciate that.
In general, $\vec x\cdot \vec y = \|\vec x\| \cdot \|\vec y\| \cos \theta$, where $\theta$ is the angle between $\vec x$ and $\vec y$. Use this fact with $\vec x = \vec a + \vec b$ and $\vec y = \vec a - \vec b$.
You already have $\|\vec x\|$ and $\|\vec y\|$. To find $\vec x \cdot \vec y$, recall the distributivity of the dot product: $$ (\vec a + \vec b) \cdot (\vec a - \vec b) = \vec a \cdot \vec a - \vec a \cdot \vec b + \vec b \cdot \vec a - \vec b \cdot \vec b = \|a\|^2 - \|b\|^2 $$