Let the angle between two nonzero vector `vecA and vecB` is `120^@` and its resultant be `vecC`.

Options

A. C must be equal to `|A-B|`
B. C must be less than `|A-B|`
C. C must be greater than `|A-B|`
D. C may be equal to `|A-B|`

Answer

To solve the problem, we need to find the magnitude of the resultant vector \( \vec{C} \) when the angle between two vectors \( \vec{A} \) and \( \vec{B} \) is \( 120^\circ \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Let the magnitudes of the vectors \( \vec{A} \) and \( \vec{B} \) be \( A \) and \( B \) respectively.
- The angle \( \theta \) between the vectors is \( 120^\circ \).

2. **Use the Formula for the Magnitude of the Resultant**:
The formula for the magnitude of the resultant vector \( \vec{C} \) when two vectors \( \vec{A} \) and \( \vec{B} \) are given by:
\[ |\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]
Here, \( \theta = 120^\circ \).

3. **Calculate \( \cos(120^\circ) \)**:
The cosine of \( 120^\circ \) is:
\[ \cos(120^\circ) = -\frac{1}{2} \]

4. **Substitute the Value of \( \cos(120^\circ) \) into the Formula**:
Now, substituting \( \cos(120^\circ) \) into the resultant formula:
\[ |\vec{C}| = \sqrt{A^2 + B^2 + 2AB \left(-\frac{1}{2}\right)} \]
Simplifying this gives:
\[ |\vec{C}| = \sqrt{A^2 + B^2 - AB} \]

5. **Final Expression for the Magnitude of the Resultant**:
Therefore, the magnitude of the resultant vector \( \vec{C} \) is:
\[ |\vec{C}| = \sqrt{A^2 + B^2 - AB} \]

### Conclusion:
The magnitude of the resultant vector \( \vec{C} \) when the angle between vectors \( \vec{A} \) and \( \vec{B} \) is \( 120^\circ \) is given by:
\[ |\vec{C}| = \sqrt{A^2 + B^2 - AB} \]