JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
Let a vector $\vec a$ has magnitude $9$. Let a vector $\vec b$ be such that for every $(x,y)\in\mathbf{R}\times\mathbf{R}-\{(0,0)\}$, the vector $(x\vec a+y\vec b)$ is perpendicular to the vector $(6y\vec a-18x\vec b)$. Then the value of $|\vec a\times\vec b|$ is equal to :
Vector Algebra
Solution
$(x\vec a+y\vec b)\cdot(6y\vec a-18x\vec b)=6xy|\vec a|^2+(6y^2-18x^2)\,\vec a\cdot\vec b-18xy|\vec b|^2=0$ for all $x,y$. Coefficients of $x^2$ and $y^2$: $\vec a\cdot\vec b=0$. Coefficient of $xy$: $|\vec a|^2=3|\vec b|^2\Rightarrow|\vec b|^2=27$. $\vec a\perp\vec b$: $|\vec a\times\vec b|=9\cdot3\sqrt3=27\sqrt3$. Option (B).