Question

If \( \vec{a} \) and \( \vec{b} \) are non-collinear unit vectors and \( |\vec{a} + \vec{b}|^2 = 3 \), then \( (3\vec{a} + 2\vec{b}) \cdot (3\vec{a} - \vec{b}) \) is equal to:

• A. \( \frac{32}{3} \)
• B. \( \frac{17}{2} \)
• C. 15
• D. 7
• E. \( \frac{17}{4} \)

Hint:

Use the formula for the square of the magnitude of a vector and the distributive property of the dot product for solving vector problems.

Answer 1

The Correct Option is B

Step 1: We are given that \( |\vec{a} + \vec{b}|^2 = 3 \). 
Expanding this expression: \[ |\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}. \] Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( \vec{a} \cdot \vec{a} = 1 \) and \( \vec{b} \cdot \vec{b} = 1 \). Therefore: \[ |\vec{a} + \vec{b}|^2 = 1 + 2 \vec{a} \cdot \vec{b} + 1 = 3. \] Simplifying: \[ 2 \vec{a} \cdot \vec{b} + 2 = 3 \quad \Rightarrow \quad 2 \vec{a} \cdot \vec{b} = 1 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = \frac{1}{2}. \] Step 2: Now, we need to calculate \( (3 \vec{a} + 2 \vec{b}) \cdot (3 \vec{a} - \vec{b}) \). 
Using the distributive property of the dot product: \[ (3 \vec{a} + 2 \vec{b}) \cdot (3 \vec{a} - \vec{b}) = 3 \vec{a} \cdot 3 \vec{a} - 3 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot 3 \vec{a} - 2 \vec{b} \cdot \vec{b}. \] 
Simplifying each term: \[ = 9 \vec{a} \cdot \vec{a} - 3 \vec{a} \cdot \vec{b} + 6 \vec{a} \cdot \vec{b} - 2 \vec{b} \cdot \vec{b}. \] Using \( \vec{a} \cdot \vec{a} = 1 \), \( \vec{b} \cdot \vec{b} = 1 \), and \( \vec{a} \cdot \vec{b} = \frac{1}{2} \), we get: \[ = 9(1) - 3\left(\frac{1}{2}\right) + 6\left(\frac{1}{2}\right) - 2(1) = 9 - \frac{3}{2} + 3 - 2. \] 
Simplifying further: \[ = 9 + 3 - 2 - \frac{3}{2} = 10 - \frac{3}{2} = \frac{20}{2} - \frac{3}{2} = \frac{17}{2}. \]