Let the angle $\theta, 0 \lt \theta \lt \frac{\pi}{2}$ between two unit vectors $\hat{\mathrm{a}}$ and…

Let the angle $\theta, 0 \lt \theta \lt \frac{\pi}{2}$ between two unit vectors $\hat{\mathrm{a}}$ and $\hat{\mathrm{b}}$ be $\sin ^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{a}}+6 \hat{\mathrm{~b}}+9(\hat{\mathrm{a}} \times \hat{\mathrm{b}}),$ then the value of $9(\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \hat{\mathrm{b}})$ is
  1. 31
  2. 27
  3. 29
  4. 24

Solution

$\begin{aligned}
& \overrightarrow{\mathrm{c}}=3 \overrightarrow{\mathrm{a}}+6 \overrightarrow{\mathrm{~b}}+9(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ & \sin ^{-1}\left(\frac{\sqrt{65}}{9}\right) \Rightarrow \sin \theta=\frac{\sqrt{65}}{9} \Rightarrow \cos \theta=\frac{4}{9} \\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=3|\overrightarrow{\mathrm{a}}|^2+6 \vec{a} \cdot \vec{b}=3+\frac{6 \cdot 4}{9}=\frac{51}{9} \\ & \overrightarrow{\mathrm{c}} \cdot \vec{a}=3 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}+6|\overrightarrow{\mathrm{~b}}|^2=\frac{3 \cdot 4}{9}+6=\frac{22}{3} \\ & \therefore 9(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-3(\overrightarrow{\mathrm{c}} \cdot \vec{b})=51-22=29
\end{aligned}$ ,

Asked in: JEE Main 2025 (07 Apr Shift 1)

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