If `vec A+vec B=vec C` and the angle between `vec A` and `vec B` is `120^(@)`, then the magnitude of `vec C`

Options

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

Answer

To find the magnitude of vector \( \vec{C} \) given that \( \vec{A} + \vec{B} = \vec{C} \) and the angle between \( \vec{A} \) and \( \vec{B} \) is \( 120^\circ \), we can use the formula for the magnitude of the resultant of two vectors.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have two vectors \( \vec{A} \) and \( \vec{B} \).
- The angle \( \theta \) between \( \vec{A} \) and \( \vec{B} \) 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:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos(\theta)} \]
- Here, \( |\vec{A}| \) is the magnitude of vector \( \vec{A} \), \( |\vec{B}| \) is the magnitude of vector \( \vec{B} \), and \( \theta = 120^\circ \).

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

4. **Substitute the Values into the Formula**:
- Now substituting \( \cos(120^\circ) \) into the formula:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \left(-\frac{1}{2}\right)} \]
- This simplifies to:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - |\vec{A}| |\vec{B}|} \]

5. **Final Expression**:
- Thus, the magnitude of vector \( \vec{C} \) is given by:
\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - |\vec{A}| |\vec{B}|} \]