JEE Main 2023 (Online) 13th April Evening ShiftQuestion 1
Let $|\vec{a}|=2$, $|\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $\left|(\vec{a}+2\vec{b})\times(2\vec{a}-3\vec{b})\right|^{2}$ is equal to
Vector Algebra
Solution
$(\vec a+2\vec b)\times(2\vec a-3\vec b)=-3\,\vec a\times\vec b+4\,\vec b\times\vec a=-7\,\vec a\times\vec b$. $|\vec a\times\vec b|^2=a^2b^2-(\vec a\cdot\vec b)^2=36-18=18$. Result $=49\cdot18=882$. Option (B).