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

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