Jee Main 2024 31 01 2024 Shift1 - Question18

Answer: 167

Solution:

$|\vec{c}|^{2}=9(|\vec{a} \times \vec{b}|)^{2}+|\vec{b}|^{2}$

$|\vec{c}|^{2}=9(16-4)+16$

$$ {\because|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} } $$

$|\vec{C}|^{2}=124$

$|\vec{c}|=\sqrt{124}$

$\vec{c}=(3(\vec{a} \times \vec{b}))-\vec{b}$

$\vec{c} \cdot \vec{b}=-|\vec{b}|^{2}=-16$

$4 \times \sqrt{124} \cos \alpha=-16$

$\cos \alpha=\frac{-4}{\sqrt{124}}=\frac{-2}{\sqrt{31}}$

$\sin \alpha=\sqrt{1-\frac{4}{31}}$

$\sin \alpha=\sqrt{\frac{27}{31}}$

Then, $192 \sin ^{2} \alpha=192 \times \frac{27}{31}$

$\approx 167.2$