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How to find $|\vec{a} \times \vec{b}|$, given that $\vec{a}=3\vec{p}+\vec{q}$, $\quad\vec{b}=\vec{p}-2\vec{q}$,$\quad|\vec{p}|=4$,$\quad|\vec{q}|=1$ and $\angle(\vec{p};\vec{q})= {{\pi} \over {4}}$?

I can extract $\vec{a}\cdot\vec{b}$ from first two equations:

$\vec{a}\cdot\vec{b}=3\vec{p}^2-6 \vec{p}\cdot \vec{q}+ \vec{q}\cdot \vec{p} -2\vec{q}^2=3|\vec{p}|^2-5|\vec{p}||\vec{q}|\cos{\angle(\vec{p};\vec{q})}-2|\vec{q}|^2=46-10\sqrt{2}$

I also know that $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin{\angle(\vec{a};\vec{b})}$

If I would somehow find out $\cos{\angle(\vec{a};\vec{b})}$, then I would be able to find $|\vec{a}||\vec{b}|$ and therefore would be able to express $|\vec{a}\times\vec{b}|$.

Is my approach right, to this problem and how would I find $\cos{\angle(\vec{a};\vec{b})}$?

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    $\begingroup$ I would try to calculate $\vec{a}\times\vec{b}$ in terms of $\vec{p}\times\vec{q}$. Some useful relations are: $\vec{p}\times\vec{p} = 0$ and $\vec{p}\times\vec{q} = -\vec{q}\times\vec{p} $ $\endgroup$
    – Winther
    Feb 3 '15 at 15:21
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$\begin{align} a\times b &= (3p+q)\times (p-2q)\\ &= 3p \times p -3p \times 2q +q\times p -q \times 2q \\ &= 0-6p \times q -p\times q +0 \\ &=-7p\times q \end{align}$

Now $|p|.|q|.sin\angle(p;q) = 4\cdot\frac{1}{\sqrt{2}}$

$\therefore |a\times b| = 14\sqrt{2}$

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