Given $\Vert \vec{u} \Vert$ and $\Vert \vec{v} \Vert$ and $\angle 120^\circ$ find volume with sides $\vec{u} \times \vec{v}$, $\vec{u}$ and $\vec{v}$

Question

I am given the following problem:

Knowing that $\Vert \vec{u} \Vert = 3$ and $\Vert \vec{v} \Vert = 4$ and also $\angle (\vec{u}, \vec{v}) = 120^\circ$ find the volume of the parallelepiped with sides $\vec{u} \times \vec{v}$, $\vec{u}$ and $\vec{v}$.

What I tried (and I am not sure if it works) is

$$ V = \Vert (\vec{u} \times \vec{v}) \cdot \vec{u} \times \vec{v} \Vert = \Vert (\vec{u} \times \vec{v}) \cdot \vec{u} + (\vec{u} \times \vec{v}) \cdot \vec{v} \Vert = \Vert \vec{u} \cdot (\vec{u} \times \vec{v}) + \vec{v} \cdot (\vec{u} \times \vec{v}) \Vert $$

Did I make a mistake somewhere?

Textbook's answer: $108 \ u.v.$

Answer 1

The "base" of this shape is a parallelogram with sides $u$ and $v$, and thus has area $||u\times v|| = ||u||\cdot||v||\sin(120) = 6\sqrt{3}$. This number is also the height of the parallelepided, since the vector $u\times v$ is perpendicular to $u$ and $v$, and so the volume is just this height times the area of the base, which is $(6\sqrt{3})^2 = 108$. You had the right formula, you just needed to compute it correctly.

Answer 2

$$V = \Vert (\vec{u} \times \vec{v}) \cdot (\vec{u} \times \vec{v} )\Vert =\Vert (\vec{u} \times \vec{v})\Vert ^2=\Vert u\Vert^2\Vert v\Vert^2\sin^2\left(\frac{2\pi}{3}\right)=108$$