MCQ
Let $|\vec{a}|=2,|\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to
- A$482$
- B$441$
- C$841$
- ✓$882$
$|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$
$|-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2$
$|-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2$
$|-7 \vec{a} \times \vec{b}|^2$
$\left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2$
$49 \times 4 \times 9 \times \frac{1}{2}=882$
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