Question

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $|\vec{a}|=4,|\vec{b}|=3\theta \in \left(\frac{\pi }{4},\frac{\pi }{3}\right)$. Then $|(\vec{a}-\vec{b})\times (\vec{a}+\vec{b}){|}^2+4(\vec{a}\cdot \vec{b}{)}^2$ is equal to _________

Answer 1

Answer: 576

$|\vec{a}|=4,|\vec{b}|=3\theta \in \left(\frac{\pi }{4},\frac{\pi }{3}\right)$
$|(\vec{a}-\vec{b})\times (\vec{a}+\vec{b}){|}^2+4(\vec{a}\cdot \vec{b}{)}^2$
$|\vec{a}\times \vec{b}-\vec{b}\times \vec{a}{|}^2+4a^2{b}^2{\cos }^2\theta$
$2|\vec{a}\times \vec{b}{|}^2+4a^2{b}^2{\cos }^2\theta$
$4a^2{b}^2{\sin }^2\theta +4a^2{b}^2{\cos }^2\theta$
$4a^2{b}^2=4\times 16\times 9=576$