JEE Main 2022 (Online) 24th June Morning ShiftQuestion 1
Let $\hat{a},\hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{\pi}{12}$, and $\hat{b}=\vec{c}+2(\vec{c}\times\hat{a})$, then $|6\vec{c}|^{2}$ is equal to :
Vector Algebra
Solution
$|\vec b|^2=|\vec c|^2+4|\vec c\times\hat a|^2+4\vec c\cdot(\vec c\times\hat a)=|\vec c|^2(1+4\sin^2\tfrac{\pi}{12})=1$. $\sin^2\tfrac{\pi}{12}=\frac{2-\sqrt3}{4}$, so $|\vec c|^2=\frac{1}{3-\sqrt3}$. $|6\vec c|^2=\frac{36}{3-\sqrt3}=\frac{36(3+\sqrt3)}{6}=6(3+\sqrt3)$. Answer: (C).