Two vectors `A` and `B` are such that `A+B = C` and `A^2 +B^2 = C^2`. If `theta` is the angle between positive direction of `A` and `B`, then the correct statement is

Options

A. `theta = pi`
B. `theta = (2pi)/(3)`
C. `theta = 0`
D. `theta = (pi)/(2)`

Explanation

To solve the problem, we need to analyze the given equations involving the vectors \( A \), \( B \), and \( C \).

### Step-by-Step Solution:

1. **Understand the Given Equations**:
We have two equations:
\[ A + B = C \]
\[ A^2 + B^2 = C^2 \]
Here, \( A \), \( B \), and \( C \) are vectors.

2. **Interpret the First Equation**:
The first equation \( A + B = C \) implies that vector \( C \) is the resultant of vectors \( A \) and \( B \). This suggests that the vectors \( A \) and \( B \) are being added head-to-tail to form \( C \).

3. **Interpret the Second Equation**:
The second equation \( A^2 + B^2 = C^2 \) can be interpreted in terms of magnitudes:
\[ |A|^2 + |B|^2 = |C|^2 \]
This is reminiscent of the Pythagorean theorem, which holds true when the vectors \( A \) and \( B \) are perpendicular to each other.

4. **Using the Law of Cosines**:
The law of cosines states:
\[ C^2 = A^2 + B^2 + 2AB \cos(\theta) \]
where \( \theta \) is the angle between vectors \( A \) and \( B \).

5. **Equating the Two Expressions for \( C^2 \)**:
Since we have \( A^2 + B^2 = C^2 \), we can substitute this into the law of cosines:
\[ A^2 + B^2 = A^2 + B^2 + 2AB \cos(\theta) \]
This simplifies to:
\[ 0 = 2AB \cos(\theta) \]

6. **Analyzing the Result**:
For the equation \( 0 = 2AB \cos(\theta) \) to hold true, either \( A \) or \( B \) must be zero, or \( \cos(\theta) = 0 \). The latter implies:
\[ \theta = 90^\circ \quad \text{(or } \theta = \frac{\pi}{2} \text{ radians)} \]
This means that vectors \( A \) and \( B \) are perpendicular to each other.

7. **Conclusion**:
Therefore, the angle \( \theta \) between the positive direction of \( A \) and \( B \) is \( 90^\circ \).

### Final Answer:
The correct statement is that \( \theta = 90^\circ \) (or \( \theta = \frac{\pi}{2} \) radians).