The two vectors `vecA` and `vecB` are drawn from a common point and `vecC=vecA+vecB`. Then, the angle between `vecA` and `vecB` is

Options

A. `90^(@)` if `C^(2)gtA^(2)+B^(2)`
B. greater than `90^(@)` if `C^(2)ltA^(2)+B^(2)`
C. greater than `90^(@)` if `C^(2)gtA^(2)+B^(2)`
D. less than `90^(@)` if `C^(2)gtA^(2)+B^(2)`

Answer

To find the angle between the two vectors \(\vec{A}\) and \(\vec{B}\), we can use the law of cosines, which relates the magnitudes of the vectors and the angle between them.

### Step-by-Step Solution:

1. **Understanding Vector Addition**:
We have two vectors \(\vec{A}\) and \(\vec{B}\) drawn from a common point. The resultant vector \(\vec{C}\) is given by:
\[ \vec{C} = \vec{A} + \vec{B} \]

2. **Using the Law of Cosines**:
The magnitude of the resultant vector \(\vec{C}\) can be expressed using the law of cosines:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos(\theta) \]
where \(\theta\) is the angle between vectors \(\vec{A}\) and \(\vec{B}\).

3. **Rearranging the Equation**:
We can rearrange the equation to isolate \(\cos(\theta)\):
\[ \cos(\theta) = \frac{|\vec{C}|^2 - |\vec{A}|^2 - |\vec{B}|^2}{2 |\vec{A}| |\vec{B}|} \]

4. **Analyzing the Angles**:
- If \(\theta = 90^\circ\): \(\cos(90^\circ) = 0\) implies \( |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 \).
- If \(\theta < 90^\circ\): \(\cos(\theta) > 0\) implies \( |\vec{C}|^2 > |\vec{A}|^2 + |\vec{B}|^2 \).
- If \(\theta > 90^\circ\): \(\cos(\theta) < 0\) implies \( |\vec{C}|^2 < |\vec{A}|^2 + |\vec{B}|^2 \).

5. **Conclusion**:
Based on the conditions, we can conclude that:
- If \( |\vec{C}|^2 > |\vec{A}|^2 + |\vec{B}|^2 \), then the angle \(\theta\) is less than \(90^\circ\).
- If \( |\vec{C}|^2 < |\vec{A}|^2 + |\vec{B}|^2 \), then the angle \(\theta\) is greater than \(90^\circ\).