JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 2)
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 :
441
482
841
882
Explanation
$$
\begin{aligned}
& |\vec{a}|=2 \\\\
& |\vec{b}|=3 \\\\
& \vec{a} \cdot \vec{b}=\frac{\pi}{4}
\end{aligned}
$$
$$ \begin{aligned} & |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\ & = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\ & = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\ & = |-7 \vec{a} \times \vec{b}|^2 \\\\ & = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2 \end{aligned} $$
= $$ 49 \times 4 \times 9 \times \frac{1}{2}=882 $$
$$ \begin{aligned} & |(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 \\\\ & = |-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2 \\\\ & = |-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2 \\\\ & = |-7 \vec{a} \times \vec{b}|^2 \\\\ & = \left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2 \end{aligned} $$
= $$ 49 \times 4 \times 9 \times \frac{1}{2}=882 $$
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