Question

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a}\times \vec{b}|=\sqrt{48}$. Then $(\vec{a}\cdot \vec{b}{)}^2$ is equal to

Answer 1

Answer: 36

$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}|\vec{a}\times \vec{b}|=\sqrt{48}$
$|\vec{a}\times \vec{b}{|}^2+|\vec{a}\cdot \vec{b}{|}^2=|\vec{a}{|}^2\times |\vec{b}{|}^2$
$\Rightarrow (\vec{a}\cdot \vec{b}{)}^2=84-48=36$