JEE Main 2025 Apr 7 Shift 1, Mathematics Q19: Scalar (or Dot) Product of Two Vectors

Step 1: Determine $\cos \theta$.

$\sin \theta =\frac{\sqrt{65}}{9}\Rightarrow \cos \theta =\sqrt{1-\frac{65}{81}}=\sqrt{\frac{16}{81}}=\frac{4}{9}$

So $\hat{a}\cdot \hat{b}=\frac{4}{9}$.

Step 2: Compute the two required dot products. Recall $\hat{a}$ and $\hat{b}$ are unit vectors and $(\hat{a}\times \hat{b})$ is perpendicular to both, so it contributes 0 in every dot product below.

$\vec{c}\cdot \hat{a}=3(\hat{a}\cdot \hat{a})+6(\hat{a}\cdot \hat{b})+9(\hat{a}\times \hat{b})\cdot \hat{a}=3+6\cdot \frac{4}{9}+0=3+\frac{24}{9}=3+\frac{8}{3}=\frac{17}{3}$

$\vec{c}\cdot \hat{b}=3(\hat{a}\cdot \hat{b})+6(\hat{b}\cdot \hat{b})+0=3\cdot \frac{4}{9}+6=\frac{4}{3}+6=\frac{22}{3}$

Step 3: Assemble the target expression.

$9(\vec{c}\cdot \hat{a})-3(\vec{c}\cdot \hat{b})=9\cdot \frac{17}{3}-3\cdot \frac{22}{3}=51-22=29$

$\therefore$ Option (3).