Evaluate $$|(2\vec{a}+3\vec{b})\times(3\vec{a}+2\vec{b})|$$ when $|\vec{a}|=2,\space|\vec{b}|=3$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$.
I've got algebra lectures years ago and don't remember the algorithm. Can anybody help me?
HINTS:
The cross product is distributive, i.e. ${\bf a} \times ({\bf b}+{\bf c}) = ({\bf a} \times {\bf b}) + ({\bf a} \times {\bf c})$. A similar result is true for $({\bf a} + {\bf b})\times{\bf c}$.
The cross product is anti-commutative, i.e. ${\bf a} \times {\bf b} = -({\bf b} \times {\bf a})$.
The magnitude of the cross product is given by $\|{\bf a} \times {\bf b}\| = \|{\bf a}\|\|{\bf b}\| \left|\sin\theta \right|$ where $\theta$ is the angle between ${\bf a}$ and ${\bf b}$ in the plane spanned by ${\bf a}$ and ${\bf b}$.