If `vec(A) + vec(B) = vec(C )` and `A + B = C`, then the angle between `vec(A)` and `vec(B)` is :

Options

A. 0
B. `(pi)/(4)`
C. `(pi)/(2)`
D. `pi`

Answer

To solve the problem, we need to find the angle between the vectors \(\vec{A}\) and \(\vec{B}\) given the equations \(\vec{A} + \vec{B} = \vec{C}\) and \(A + B = C\).

### Step-by-Step Solution:

1. **Understand the Given Equations:**
- We have two equations:
\[ \vec{A} + \vec{B} = \vec{C} \]
\[ A + B = C \]
- Here, \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) are vectors, while \(A\), \(B\), and \(C\) are their magnitudes.

2. **Magnitude of the Vectors:**
- The magnitude of the vector sum can be expressed using the cosine of the angle \(\theta\) between the two vectors:
\[ |\vec{A} + \vec{B}| = |\vec{C}| \]
- In terms of magnitudes, this can be written as:
\[ |A + B| = C \]

3. **Using the Cosine Rule for Vectors:**
- The formula for the magnitude of the sum of two vectors is:
\[ |\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \]
- Substituting the magnitudes, we have:
\[ C^2 = A^2 + B^2 + 2AB \cos \theta \]

4. **Setting Up the Equation:**
- From the first equation \(A + B = C\), we can square both sides:
\[ (A + B)^2 = C^2 \]
- Expanding the left side gives:
\[ A^2 + 2AB + B^2 = C^2 \]

5. **Equating the Two Expressions for \(C^2\):**
- We now have two expressions for \(C^2\):
1. \(C^2 = A^2 + B^2 + 2AB \cos \theta\)
2. \(C^2 = A^2 + 2AB + B^2\)
- Setting them equal to each other:
\[ A^2 + B^2 + 2AB \cos \theta = A^2 + 2AB + B^2 \]

6. **Simplifying the Equation:**
- Cancel \(A^2 + B^2\) from both sides:
\[ 2AB \cos \theta = 2AB \]
- Dividing both sides by \(2AB\) (assuming \(A\) and \(B\) are not zero):
\[ \cos \theta = 1 \]

7. **Finding the Angle:**
- The equation \(\cos \theta = 1\) implies that:
\[ \theta = 0^\circ \]
- This means that the vectors \(\vec{A}\) and \(\vec{B}\) are in the same direction.

### Final Answer:
The angle between \(\vec{A}\) and \(\vec{B}\) is \(0^\circ\).