Key Concepts and Formulas

1. Dot Product of Unit Vectors: For two unit vectors $\hat{a}$ and $\hat{b}$, $\hat{a} \cdot \hat{b} = \cos\theta$, where $\theta$ is the angle between them. Also, $\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1$.
2. Orthogonality of Cross Product: The cross product $\hat{a} \times \hat{b}$ is orthogonal to both $\hat{a}$ and $\hat{b}$. Thus, $(\hat{a} \times \hat{b}) \cdot \hat{a} = 0$ and $(\hat{a} \times \hat{b}) \cdot \hat{b} = 0$.
3. Trigonometric Identity: $\sin^2\theta + \cos^2\theta = 1$.

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Step-by-Step Solution

We are given a vector $\vec{c} = 3 \hat{a} + 6 \hat{b} + 9(\hat{a} \times \hat{b})$, where $\hat{a}$ and $\hat{b}$ are unit vectors. The angle $\theta$ between them satisfies $\sin\theta = \frac{\sqrt{65}}{9}$ and $0 < \theta < \frac{\pi}{2}$. We need to find the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$.

Step 1: Calculate $\vec{c} \cdot \hat{a}$ We compute the dot product of $\vec{c}$ with $\hat{a}$ using the distributive property of the dot product. $\vec{c} \cdot \hat{a} = (3 \hat{a} + 6 \hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{a}$ $\vec{c} \cdot \hat{a} = (3 \hat{a} \cdot \hat{a}) + (6 \hat{b} \cdot \hat{a}) + (9(\hat{a} \times \hat{b}) \cdot \hat{a})$ Using the properties of unit vectors and the orthogonality of the cross product:

• $3 \hat{a} \cdot \hat{a} = 3 |\hat{a}|^2 = 3(1)^2 = 3$.
• $6 \hat{b} \cdot \hat{a} = 6 (\hat{a} \cdot \hat{b})$.
• $9(\hat{a} \times \hat{b}) \cdot \hat{a} = 0$ because $\hat{a} \times \hat{b}$ is orthogonal to $\hat{a}$. Therefore, $\vec{c} \cdot \hat{a} = 3 + 6(\hat{a} \cdot \hat{b})$

Step 2: Calculate $\vec{c} \cdot \hat{b}$ Similarly, we compute the dot product of $\vec{c}$ with $\hat{b}$. $\vec{c} \cdot \hat{b} = (3 \hat{a} + 6 \hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{b}$ $\vec{c} \cdot \hat{b} = (3 \hat{a} \cdot \hat{b}) + (6 \hat{b} \cdot \hat{b}) + (9(\hat{a} \times \hat{b}) \cdot \hat{b})$ Using the properties of unit vectors and the orthogonality of the cross product:

• $3 \hat{a} \cdot \hat{b} = 3 (\hat{a} \cdot \hat{b})$.
• $6 \hat{b} \cdot \hat{b} = 6 |\hat{b}|^2 = 6(1)^2 = 6$.
• $9(\hat{a} \times \hat{b}) \cdot \hat{b} = 0$ because $\hat{a} \times \hat{b}$ is orthogonal to $\hat{b}$. Therefore, $\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6$

Step 3: Substitute into the Target Expression Now we substitute the expressions for $\vec{c} \cdot \hat{a}$ and $\vec{c} \cdot \hat{b}$ into the required expression $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$. $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9(3 + 6(\hat{a} \cdot \hat{b})) - 3(3(\hat{a} \cdot \hat{b}) + 6)$ Distribute the constants: $= (27 + 54(\hat{a} \cdot \hat{b})) - (9(\hat{a} \cdot \hat{b}) + 18)$ Combine like terms: $= 27 + 54(\hat{a} \cdot \hat{b}) - 9(\hat{a} \cdot \hat{b}) - 18$ $= (27 - 18) + (54 - 9)(\hat{a} \cdot \hat{b})$ $= 9 + 45(\hat{a} \cdot \hat{b})$

Step 4: Determine the Value of $\hat{a} \cdot \hat{b}$ We are given that $\hat{a}$ and $\hat{b}$ are unit vectors and the angle between them is $\theta$, with $\sin\theta = \frac{\sqrt{65}}{9}$ and $0 < \theta < \frac{\pi}{2}$. We know that $\hat{a} \cdot \hat{b} = \cos\theta$. Using the identity $\sin^2\theta + \cos^2\theta = 1$: $\cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{\sqrt{65}}{9}\right)^2$ $\cos^2\theta = 1 - \frac{65}{81} = \frac{81 - 65}{81} = \frac{16}{81}$ Since $0 < \theta < \frac{\pi}{2}$, $\cos\theta$ is positive. $\cos\theta = \sqrt{\frac{16}{81}} = \frac{4}{9}$ Thus, $\hat{a} \cdot \hat{b} = \frac{4}{9}$.

Step 5: Final Calculation Substitute the value of $\hat{a} \cdot \hat{b}$ into the simplified expression from Step 3: $9 + 45(\hat{a} \cdot \hat{b}) = 9 + 45 \left(\frac{4}{9}\right)$ $= 9 + \frac{45}{9} \times 4$ $= 9 + 5 \times 4$ $= 9 + 20$ $= 29$

The value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$ is 29.

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Common Mistakes & Tips

• Orthogonality Simplification: The most significant simplification comes from recognizing that the cross product term becomes zero when dotted with either $\hat{a}$ or $\hat{b}$. Failing to use this property can lead to unnecessarily complicated calculations.
• Trigonometric Sign: Always consider the quadrant of the angle $\theta$ to determine the correct sign of trigonometric functions like $\cos\theta$. In this case, $0 < \theta < \frac{\pi}{2}$ implies $\cos\theta > 0$.
• Unit Vector Properties: Remember that for unit vectors, $\hat{u} \cdot \hat{u} = 1$, which is crucial for simplifying terms like $3\hat{a} \cdot \hat{a}$.

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Summary

The problem involves calculating a linear combination of dot products of a vector $\vec{c}$ with two unit vectors $\hat{a}$ and $\hat{b}$. By strategically applying the distributive property of the dot product and the orthogonality property of the cross product, the terms involving $\hat{a} \times \hat{b}$ conveniently vanish. The remaining expression simplifies to a constant plus a multiple of $\hat{a} \cdot \hat{b}$. We then use the given sine of the angle between $\hat{a}$ and $\hat{b}$ to find the cosine, which is $\hat{a} \cdot \hat{b}$, and substitute it to obtain the final numerical value.

The final answer is $\boxed{29}$, which corresponds to option (A).