If `veca and vecb` are any two unit vectors, then find the greatest postive integer in the range of `(3|veca + vecb|)/2+2|veca-vecb|`

Explanation

To solve the problem, we need to find the greatest positive integer in the range of the expression \(\frac{3|\vec{a} + \vec{b}|}{2} + 2|\vec{a} - \vec{b}|\) given that \(\vec{a}\) and \(\vec{b}\) are unit vectors.

### Step-by-step Solution:

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

2. **Finding \(|\vec{a} + \vec{b}|\)**:
We can use the formula for the magnitude of the sum of two vectors:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \]
Substituting the values:
\[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b}) = 2 + 2(\vec{a} \cdot \vec{b}) = 2 + 2\cos\theta \]
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

Taking the square root:
\[ |\vec{a} + \vec{b}| = \sqrt{2 + 2\cos\theta} = \sqrt{2(1 + \cos\theta)} = \sqrt{4\cos^2\left(\frac{\theta}{2}\right)} = 2\cos\left(\frac{\theta}{2}\right) \]

3. **Finding \(|\vec{a} - \vec{b}|\)**:
Similarly, for the difference:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 2 - 2\cos\theta = 2(1 - \cos\theta) \]
Taking the square root:
\[ |\vec{a} - \vec{b}| = \sqrt{2(1 - \cos\theta)} = \sqrt{4\sin^2\left(\frac{\theta}{2}\right)} = 2\sin\left(\frac{\theta}{2}\right) \]

4. **Substituting into the Expression**:
Now substituting these values into the original expression:
\[ \frac{3|\vec{a} + \vec{b}|}{2} + 2|\vec{a} - \vec{b}| = \frac{3(2\cos\left(\frac{\theta}{2}\right))}{2} + 2(2\sin\left(\frac{\theta}{2}\right)} \]
Simplifying:
\[ = 3\cos\left(\frac{\theta}{2}\right) + 4\sin\left(\frac{\theta}{2}\right) \]

5. **Finding the Range**:
To find the maximum value of \(3\cos\left(\frac{\theta}{2}\right) + 4\sin\left(\frac{\theta}{2}\right)\), we can use the method of Lagrange multipliers or the Cauchy-Schwarz inequality.

Let \(x = \cos\left(\frac{\theta}{2}\right)\) and \(y = \sin\left(\frac{\theta}{2}\right)\). Then we have:
\[ x^2 + y^2 = 1 \]
The expression becomes:
\[ 3x + 4y \]
The maximum occurs when:
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]

Therefore, the maximum value of \(3\cos\left(\frac{\theta}{2}\right) + 4\sin\left(\frac{\theta}{2}\right)\) is \(5\).

6. **Conclusion**:
The greatest positive integer in the range of the expression is:
\[ \boxed{5} \]