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

<p>To solve the problem, we begin by examining the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$.</p> <p>Let's calculate:</p> <p><p>$\vec{c} \cdot \hat{a} = 3 + 6(\hat{a} \cdot \hat{b})$. </p></p> <p><p>$\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6$.</p></p> <p>We need to find the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$.</p> <p>Substituting the expressions, we get:</p> <p>$ 9(\vec{c} \cdot \hat{a}) = 9(3 + 6\hat{a} \cdot \hat{b}) = 27 + 54(\hat{a} \cdot \hat{b}) $</p> <p>$ 3(\vec{c} \cdot \hat{b}) = 3(3\hat{a} \cdot \hat{b} + 6) = 9\hat{a} \cdot \hat{b} + 18 $</p> <p>Therefore, </p> <p>$ 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = (27 + 54(\hat{a} \cdot \hat{b})) - (9\hat{a} \cdot \hat{b} + 18) $</p> <p>Simplifying,</p> <p>$ = 27 - 18 + 54(\hat{a} \cdot \hat{b}) - 9(\hat{a} \cdot \hat{b}) $</p> <p>$ = 9 + 45(\hat{a} \cdot \hat{b}) $</p> <p>Given $\sin{\theta} = \frac{\sqrt{65}}{9}$, and knowing that for unit vectors $\cos{\theta} = \hat{a} \cdot \hat{b}$, we use the identity $(\cos{\theta})^2 = 1 - (\sin{\theta})^2$:</p> <p>$ (\hat{a} \cdot \hat{b})^2 = 1 - \left(\frac{\sqrt{65}}{9}\right)^2 $</p> <p>$ = 1 - \frac{65}{81} $</p> <p>$ = \frac{16}{81} $</p> <p>Thus, $\hat{a} \cdot \hat{b} = \frac{4}{9}$.</p> <p>Substitute back into the equation:</p> <p>$ 9 + 45 \times \frac{4}{9} = 9 + 20 = 29 $</p> <p>Therefore, the value is 29.</p>