If `veca` and `vecb` are two non collinear unit vectors and `|veca+vecb|=sqrt(3)` then find the value of `(veca-vecb).(2veca+vecb)`

Options

A. `45^(@), 60^(@)`
B. `30^(@), 60^(@)`
C. `60^(@), 60^(@)`
D. `30^(@), 45^(@)`

Explanation

To solve the problem, we need to find the value of \((\vec{a} - \vec{b}) \cdot (2\vec{a} + \vec{b})\), given that \(\vec{a}\) and \(\vec{b}\) are two non-collinear unit vectors and \(|\vec{a} + \vec{b}| = \sqrt{3}\).

### Step-by-step Solution:

1. **Understanding the Magnitudes**:
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

2. **Using the Magnitude Formula**:
We know that the magnitude of the sum of two vectors can be expressed as:
\[ |\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b}} \]
Substituting the magnitudes:
\[ |\vec{a} + \vec{b}| = \sqrt{1^2 + 1^2 + 2 \vec{a} \cdot \vec{b}} = \sqrt{2 + 2 \vec{a} \cdot \vec{b}} \]
Given that \(|\vec{a} + \vec{b}| = \sqrt{3}\), we can set up the equation:
\[ \sqrt{2 + 2 \vec{a} \cdot \vec{b}} = \sqrt{3} \]

3. **Squaring Both Sides**:
Squaring both sides gives:
\[ 2 + 2 \vec{a} \cdot \vec{b} = 3 \]
Rearranging this, we find:
\[ 2 \vec{a} \cdot \vec{b} = 1 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = \frac{1}{2} \]

4. **Finding the Dot Product**:
Now we need to compute \((\vec{a} - \vec{b}) \cdot (2\vec{a} + \vec{b})\):
\[ (\vec{a} - \vec{b}) \cdot (2\vec{a} + \vec{b}) = \vec{a} \cdot (2\vec{a} + \vec{b}) - \vec{b} \cdot (2\vec{a} + \vec{b}) \]

5. **Expanding the Dot Products**:
Expanding the first term:
\[ \vec{a} \cdot (2\vec{a} + \vec{b}) = 2\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} = 2(1) + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2} \]
Expanding the second term:
\[ \vec{b} \cdot (2\vec{a} + \vec{b}) = 2\vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} = 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2 \]

6. **Combining the Results**:
Now, substituting back into our expression:
\[ (\vec{a} - \vec{b}) \cdot (2\vec{a} + \vec{b}) = \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2} \]

### Final Answer:
Thus, the value of \((\vec{a} - \vec{b}) \cdot (2\vec{a} + \vec{b})\) is:
\[ \frac{1}{2} \]