The vectors `vec A and vec B` are such that `vec A + vec B and A^(2)+B^(2) =C^(2)`. Angle between the positive directions of vectors `vec A and vec B` is equal to

Options

A. `pi` rad
B. `pi//2` rad
C. `pi//32` rad
D. `pi//4` rad

Explanation

To solve the problem, we need to find the angle between the vectors \(\vec{A}\) and \(\vec{B}\) given the conditions \(\vec{A} + \vec{B} = \vec{C}\) and \(A^2 + B^2 = C^2\).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We are given two equations:
\[ \vec{A} + \vec{B} = \vec{C} \]
\[ A^2 + B^2 = C^2 \]
Here, \(A\) and \(B\) are the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\) respectively, and \(C\) is the magnitude of vector \(\vec{C}\).

2. **Squaring the First Equation**:
We square the first equation:
\[ (\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = \vec{C} \cdot \vec{C} \]
Expanding this using the dot product:
\[ A^2 + B^2 + 2 \vec{A} \cdot \vec{B} = C^2 \]

3. **Substituting the Second Condition**:
From the second condition, we know that:
\[ A^2 + B^2 = C^2 \]
We can substitute this into our expanded equation:
\[ C^2 + 2 \vec{A} \cdot \vec{B} = C^2 \]

4. **Simplifying the Equation**:
By subtracting \(C^2\) from both sides, we have:
\[ 2 \vec{A} \cdot \vec{B} = 0 \]

5. **Interpreting the Result**:
The dot product \(\vec{A} \cdot \vec{B}\) can be expressed as:
\[ \vec{A} \cdot \vec{B} = AB \cos \theta \]
where \(\theta\) is the angle between the vectors \(\vec{A}\) and \(\vec{B}\). Therefore, we have:
\[ 2AB \cos \theta = 0 \]

6. **Finding the Angle**:
Since \(A\) and \(B\) are magnitudes and cannot be zero, we conclude:
\[ \cos \theta = 0 \]
The angle \(\theta\) that satisfies this condition is:
\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{(or 90 degrees)} \]

### Final Answer:
The angle between the positive directions of vectors \(\vec{A}\) and \(\vec{B}\) is \(\frac{\pi}{2}\) radians.