Let the angle between two non - zero vectors `vec(A)` and `vec(B)` be `120^(@)`and its resultant be `vec( C)`.

Options

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

Answer

To solve the problem regarding the resultant of two non-zero vectors \(\vec{A}\) and \(\vec{B}\) with an angle of \(120^\circ\) between them, we can follow these steps:

### Step-by-Step Solution:

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

2. **Use the Formula for the Magnitude of the Resultant:**
The magnitude of the resultant vector \(\vec{C}\) can be calculated using the formula:
\[ |\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]

3. **Substitute the Values:**
- Substitute \(\theta = 120^\circ\) into the formula. We know that \(\cos(120^\circ) = -\frac{1}{2}\).
\[ |\vec{C}| = \sqrt{A^2 + B^2 + 2AB \left(-\frac{1}{2}\right)} \]

4. **Simplify the Expression:**
- This simplifies to:
\[ |\vec{C}| = \sqrt{A^2 + B^2 - AB} \]

5. **Analyze the Resultant Magnitude:**
- We can analyze the relationship between the magnitudes of the vectors and the resultant:
\[ |\vec{C}| = \sqrt{A^2 + B^2 - AB} \]
- We can also derive inequalities based on the triangle inequality theorem. The resultant \(|\vec{C}|\) must satisfy:
\[ |\vec{C}| > |A - B| \]
- This means that the magnitude of the resultant vector is always greater than the absolute difference of the magnitudes of the two vectors.

### Conclusion:
Thus, we conclude that the magnitude of the resultant vector \(|\vec{C}|\) is given by:
\[ |\vec{C}| = \sqrt{A^2 + B^2 - AB} \]