If `vec(A) + vec(B) = vec(C )` under what condition `A^(2)+B^(2)` will be equal to `C^(2)` ?

Explanation

To solve the problem, we need to find the condition under which the equation \( A^2 + B^2 = C^2 \) holds true, given that \( \vec{A} + \vec{B} = \vec{C} \).

### Step-by-Step Solution:

1. **Start with the given vector equation:**
\[ \vec{A} + \vec{B} = \vec{C} \]

2. **Take the magnitude of both sides:**
\[ |\vec{A} + \vec{B}| = |\vec{C}| \]

3. **Use the formula for the magnitude of the sum of two vectors:**
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta} \]
where \( \theta \) is the angle between vectors \( \vec{A} \) and \( \vec{B} \).

4. **Substituting into the equation:**
\[ \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta} = |\vec{C}| \]

5. **Square both sides to eliminate the square root:**
\[ |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta = |\vec{C}|^2 \]

6. **Rearranging the equation gives us:**
\[ |\vec{A}|^2 + |\vec{B}|^2 = |\vec{C}|^2 - 2|\vec{A}||\vec{B}|\cos\theta \]

7. **To satisfy the condition \( A^2 + B^2 = C^2 \), we need:**
\[ 2|\vec{A}||\vec{B}|\cos\theta = 0 \]

8. **Since \( |\vec{A}| \) and \( |\vec{B}| \) are non-zero (assuming both vectors are not zero), the only way for the product to be zero is if:**
\[ \cos\theta = 0 \]

9. **This implies that:**
\[ \theta = \frac{\pi}{2} \]

10. **Conclusion:**
The condition under which \( A^2 + B^2 = C^2 \) is that the vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular to each other.