If `vec(a) and vec(b)` are two unit vectors inclined at an angle `60^(@)` to each other, then which one of the following is correct?

Options

A. `|vec(a)+vec(b)|lt1`
B. `|vec(a)+vec(b)|gt1`
C. `|vec(a)-vec(b)|lt1`
D. `|vec(a)-vec(b)|gt1`

Explanation

To solve the problem, we need to analyze the relationship between two unit vectors \(\vec{a}\) and \(\vec{b}\) that are inclined at an angle of \(60^\circ\) to each other. We will use the properties of vectors and the cosine rule to find the magnitude of the sum of the two vectors.

### Step-by-Step Solution:

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

2. **Using the Cosine Rule**:
The cosine rule states that for any two vectors \(\vec{u}\) and \(\vec{v}\):
\[ |\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2|\vec{u}||\vec{v}|\cos(\theta) \]
Here, \(\theta\) is the angle between the vectors. For our case:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos(60^\circ) \]

3. **Substituting Values**:
Since both vectors are unit vectors:
\[ |\vec{a}|^2 = 1, \quad |\vec{b}|^2 = 1, \quad \text{and} \quad \cos(60^\circ) = \frac{1}{2} \]
Therefore, substituting these values into the equation:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 + 2 \cdot 1 \cdot 1 \cdot \frac{1}{2} \]
Simplifying this gives:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 + 1 = 3 \]

4. **Finding the Magnitude**:
Taking the square root of both sides:
\[ |\vec{a} + \vec{b}| = \sqrt{3} \]

5. **Comparing Magnitudes**:
Now we can compare the magnitude of \(\vec{a} + \vec{b}\) with the magnitudes of the individual vectors:
- The magnitude of \(\vec{a}\) and \(\vec{b}\) is \(1\).
- The magnitude of \(\vec{a} + \vec{b}\) is \(\sqrt{3} \approx 1.732\).

6. **Conclusion**:
Since \(\sqrt{3} > 1\), we conclude that:
\[ |\vec{a} + \vec{b}| > 1 \]
Thus, the correct statement is that \(\vec{a} + \vec{b}\) is greater than \(1\).

### Final Answer:
The correct option is that \(|\vec{a} + \vec{b}| > 1\).