JEE Main 2024 (Online) 31st January Morning ShiftQuestion 1
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 $\vec{c}=\left(2\vec{a}\times\vec{b}\right)-3\vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$, then $192\sin^{2}\alpha$ is equal to $\_\_\_\_\_\_$ .
Vector Algebra
Answer$48$
Solution
$\vec a\times\vec b\perp\vec b$, so $\vec b\cdot\vec c=-3|\vec b|^2=-48$. $|\vec a\times\vec b|^2=16-4=12$, so $|\vec c|^2=4\cdot12+9\cdot16=192$. $\cos^2\alpha=\dfrac{48^2}{16\cdot192}=\dfrac34\Rightarrow\sin^2\alpha=\dfrac14$. $192\sin^2\alpha=48$.