JEE Main 2024 (Online) 5th April Evening ShiftQuestion 1
Consider three vectors $\vec{a},\vec{b},\vec{c}$. Let $|\vec{a}|=2$, $|\vec{b}|=3$ and $\vec{a}=\vec{b}\times\vec{c}$. If $\alpha\in\left[0,\dfrac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27\left|\vec{c}-\vec{a}\right|^{2}$ is equal to :
Vector Algebra
Solution
$|\vec a|=3|\vec c|\sin\alpha=2$ and $\vec a\perp\vec c$, so $|\vec c-\vec a|^2=\frac{4}{9\sin^2\alpha}+4$, least at $\alpha=\frac\pi3$: $\frac{16}{27}+4$. Hence $27|\vec c-\vec a|^2\ge16+108=124$.