Question

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:

• A. 31
• B. 27
• C. 29
• D. 24

Hint:

For problems involving vector calculations, break them down into dot products and cross products, and use the properties of unit vectors to simplify the computation.

Answer 1

The Correct Option is C

Step 1: Determine \(\cos\theta\)
Given:
\[ \theta = \sin^{-1}\left(\frac{\sqrt{65}}{9}\right) \] Thus: \[ \sin\theta = \frac{\sqrt{65}}{9} \] Using the identity \(\sin^2\theta + \cos^2\theta = 1\): \[ \cos\theta = \sqrt{1 - \left(\frac{\sqrt{65}}{9}\right)^2} = \sqrt{1 - \frac{65}{81}} = \sqrt{\frac{16}{81}} = \frac{4}{9} \] 
Step 2: Compute Dot Products
Since \(\hat{a}\) and \(\hat{b}\) are unit vectors: \[ \hat{a} \cdot \hat{b} = \cos\theta = \frac{4}{9} \] \[ \hat{a} \cdot (\hat{a} \times \hat{b}) = 0 \quad \text{(since \(\hat{a} \times \hat{b}\) is perpendicular to \(\hat{a}\))} \] \[ \hat{b} \cdot (\hat{a} \times \hat{b}) = 0 \quad \text{(since \(\hat{a} \times \hat{b}\) is perpendicular to \(\hat{b}\))} \] 
Step 3: Compute \(\vec{c} \cdot \hat{a}\) and \(\vec{c} \cdot \hat{b}\)
Using the expression for \(\vec{c}\):
\[ \vec{c} \cdot \hat{a} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{a} = 3 + 6(\hat{b} \cdot \hat{a}) + 0 = 3 + 6 \times \frac{4}{9} = 3 + \frac{24}{9} = \frac{17}{3} \] \[ \vec{c} \cdot \hat{b} = (3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})) \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6 + 0 = 3 \times \frac{4}{9} + 6 = \frac{12}{9} + 6 = \frac{22}{3} \] 
Step 4: Compute the Required Expression
Now calculate:
\[ 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9 \times \frac{17}{3} - 3 \times \frac{22}{3} = 51 - 22 = 29 \] 
Step 5: Match with Options
The result is 29, which corresponds to option (3).