JEE Main 2025 (Online) 7th April Morning ShiftQuestion 1
Let the angle $\theta$, $0<\theta<\frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c}=3\hat{a}+6\hat{b}+9(\hat{a}\times\hat{b})$, then the value of $9(\vec{c}\cdot\hat{a})-3(\vec{c}\cdot\hat{b})$ is
Vector Algebra
Solution
$\cos\theta=\sqrt{1-65/81}=\tfrac49$. $\vec c\cdot\hat a=3+6\cos\theta=\tfrac{17}{3}$ and $\vec c\cdot\hat b=3\cos\theta+6=\tfrac{22}{3}$, since $(\hat a\times\hat b)$ is perpendicular to both. $9\cdot\tfrac{17}{3}-3\cdot\tfrac{22}{3}=51-22=29$. Answer: (C).