JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 23)
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 ________.
Answer
48
Explanation
$$\begin{aligned}
& \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{b}}-3|\mathrm{b}|^2 \\
& |\mathrm{~b}||c| \cos \alpha=-3|\mathrm{~b}|^2 \\
& |\mathrm{c}| \cos \alpha=-12 \text {, as }|\mathrm{b}|=4 \\
& \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=2 \\
& \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \\
& |c|^2=|(2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-3 \overrightarrow{\mathrm{b}}|^2 \\
& =64 \times \frac{3}{4}+144=192 \\
& |c|^2 \cos ^2 \alpha=144 \\
& 192 \cos ^2 \alpha=144 \\
& 192 \sin ^2 \alpha=48
\end{aligned}$$
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