Find the angle between vectors Parallelogram constructed on the vectors $a=5p-2q$, $b=3p+2q$, $|p|=2$, $|q|=3$, $(p\wedge q)=120^{\circ}$.
$\sin(a\wedge b)=???$
 A: Hint
You can use Dot Product.
1) $a\cdot b=(5p-2q)\cdot (3p+2q)=15|p|^2-4|q|^2+4p\cdot q$
2) $p\cdot q=|p||q|\cos 120°$
Now you know $a\cdot b$.
3) $a\cdot b=|a||b|\cos x$ where $x$ is the angle between $a$ and $b$.
But
4) $|a|^2=a\cdot a=(5p-2q)(5p-2q)$ and $|b|^2=b\cdot b=(3p+2q)(3p+2q)$
Doing the above calculations you can find $\cos x$ and then $\sin x$.
Can you finish?
A: Hint: The cross product of two vectors in three dimensional space has the property that 
$$
\|\vec{v}\times\vec{w}\|=\|\vec{v}\|\|\vec{w}\|\sin(\theta)
$$
where $\theta$ is the angle between $\vec{v}$ and $\vec{w}$.
In your case, let $\varphi$ be the angle between $a$ and $b$.  Then, 
$$
\sin(\varphi)=\frac{\|a\times b\|}{\|a\|\|b\|}.
$$
Observe that $a\times b=(5p-2q)\times(3p+2q)=15p\times p-6q\times p+10p\times q-4q\times q.$  Since the cross product of a vector with itself is $0$ and rearranging the factors in a cross product changes the sign, we have that the cross product is
$$
a\times b=16p\times q.
$$
Therefore, 
$$
\|a\times b\|=16\|p\times q\|=16\|p\|\|q\|\sin(120^\circ)=96\cdot\frac{\sqrt{3}}{2}=48\sqrt{3}.
$$
Now, we want to calculate $\|a\|$.  Using the dot product, 
$$
\|a\|^2=a\cdot a=(5p-2q)\cdot(5p-2q)=25\|q\|^2-20p\cdot q+4\|q\|^2
$$
Substituting, $\|a\|^2=241-20p\cdot q$.  Finally, the dot product of two vectors is $p\cdot q=\|p\|\|q\|\cos(120^\circ)=-3$.  Therefore, $\|a\|^2=301$.
I'll let you continue from here.  
Also note, as an alternative, you can just make up two vectors for $p$ and $q$ for example $p=(2,0)$ and $q=\left(3\cos(120^\circ),3\sin(120^\circ)\right)$ and then computing $a$ and $b$ and compute from there.
