$\lvert \vec{a} \rvert = 2$, $\lvert \vec{b} \rvert = 3$, and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$. What is the area of the parallelogram with adjacent sides $2\vec{a} + \vec{b}$ and $\vec{a} + 3\vec{b}$?

Explanation

1. 01

   $(2\vec{a} + \vec{b}) \times (\vec{a} + 3\vec{b}) = 2\vec{a} \times \vec{a} + 6\vec{a} \times \vec{b} + \vec{b} \times \vec{a} + 3\vec{b} \times \vec{b}$.

2. 02

   $\vec{a} \times \vec{a} = \vec{b} \times \vec{b} = \vec{0}$ and $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$, so it is $5\vec{a} \times \vec{b}$.

3. 03

   $\lvert \vec{a} \times \vec{b} \rvert = 2 \times 3 \times \sin\frac{\pi}{6} = 3$, so the area is $5 \times 3 = 15$.