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Let \(\vec{a}\) and \(\vec{b}\) be two vectors such that \(|\vec{a}|=1,|\vec{b}|=4\) and \(\vec{a} \cdot \vec{b}=2\). If \(\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}\) and the angle between \(\vec{b}\) and \(\vec{c}\) is \(\alpha\), then \(192 \sin ^{2} \alpha\) is equal to _____.

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Correct answer: 48

\(\vec{b} \cdot \vec{c}=(2 \vec{a} \times \vec{b}) \cdot \vec{b}-3|b|^{2}\)

\(|\mathrm{b}||\mathrm{c}| \cos \alpha=-3|\mathrm{~b}|^{2}\)

\(|c| \cos \alpha=-12,\text{ as } |b|=4\)

\(\vec{{a}} \cdot \vec{{b}}=2\)

\(\cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\)

\(|c|^{2}=|(2 \vec{a} \times \vec{b})-3 \vec{b}|^{2}\)

\(=64 \times \frac{3}{4}+144=192\)

\(|c|^{2} \cos ^{2} \alpha=144\)

\(192 \cos ^{2} \alpha=144\)

\(192 \sin ^{2} \alpha=48\)

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