Find area of parallelogram that is formed using vectors $\vec{a}$ and $\vec{b}$. It is also given that
$\vec{a} = 3\vec{p}+\vec{q}$, $\vec{b} = \vec{p}-2\vec{q}$, $|\vec{p}|=4$, $|\vec{q}|=1$, $<(p,q) = \pi / 4$
I only know that Area Of Parallelogram is equal to $|\vec{a}||\vec{b}|\cdot sin(\alpha)$, where $\alpha$ is angle between two vectors.
Best Answer
Hint: It is also equal to
$$||\vec{a}\times\vec{b}||$$
where $\times$ means the cross product.
Find cross product of $\vec{a},\vec{b}$ in terms of cross product of $\vec{p},\vec{q}$, using distributive law. Remember the fact that
$$\vec{p}\times \vec{p}=0\\ \vec{p}\times \vec{q}=-\vec{q}\times \vec{p}$$
Then when you will need to find the norm of $\vec{p}\times \vec{q}$, which is $|p| |q| \sin{\alpha}$.