Question

If $\left|\vec{a}\right|=4, \left|\vec{b}\right|=2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\pi/6$ then $\left(\vec{a}\times\vec{b}\right)^{2}$ is equal to _______.

Answer 1

We have
$\vec{a}\cdot\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right| cos \frac{\pi}{6}$
$=4\times2\times \frac{\sqrt{3}}{2}=4\sqrt{3} $
Now, $\left(\vec{a}\times\vec{b}\right)^{2} +\left(\vec{a}\cdot\vec{b}\right)^{2}=a^{2}b^{2}$
$\Rightarrow \left(\vec{a}\times\vec{b}\right)^{2}=48=16\times4$
$\Rightarrow \left(\vec{a}\times\vec{b}\right)^{2}=16$