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QJEEM24Ma5192: JEE Main 2024 Physics - Class 11 JEEM24Ma5192 Solution: Let and be two vectors such that , and . If and the angle…

Let a⃗\vec{a} and b⃗\vec{b} be two vectors such that ∣a⃗∣=1,∣b⃗∣=4|\vec{a}|=1,|\vec{b}|=4 , and a⃗⋅b⃗=2\vec{a} \cdot \vec{b}=2 . If c⃗=(2a⃗×b⃗)−3b⃗\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b} and the angle between b⃗\vec{b} and c⃗\vec{c} is α\alpha , then 192sin⁡2α192 \sin ^2 \alpha is equal to ________.

Step-by-Step Verified Solution

amp;b→⋅c→=(2a→×b→)⋅b→−3∣b∣2amp;∣ b∣∣c∣cos⁡α=−3∣ b∣2amp;∣c∣cos⁡α=−12, as ∣b∣=4amp;a→⋅b→=2amp;cos⁡θ=12⇒θ=π3amp;∣c∣2=∣(2a→×b→)−3b→∣2amp;=64×34+144=192amp;∣c∣2cos⁡2α=144amp;192cos⁡2α=144amp;192sin⁡2α=48\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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