If `vec(a)` and `vec(b)` are perpendicular vectors, `|vec(a)+vec(b)|=3` and `|vec(a)|=5`, find the value of `|vec(b)|`.

Answer

To solve the problem, we will use the properties of vectors and the given information step by step.

### Given:
- Vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular.
- \( |\vec{a} + \vec{b}| = 3 \)
- \( |\vec{a}| = 5 \)

### To find:
- \( |\vec{b}| \)

### Step-by-step Solution:

1. **Use the formula for the magnitude of the sum of two vectors**:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos \theta \]
Since \( \vec{a} \) and \( \vec{b} \) are perpendicular, \( \theta = 90^\circ \) and \( \cos 90^\circ = 0 \). Therefore, the formula simplifies to:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 \]

2. **Substituting the known values**:
\[ |\vec{a} + \vec{b}|^2 = 3^2 = 9 \]
\[ |\vec{a}|^2 = 5^2 = 25 \]
Thus, we can write:
\[ 9 = 25 + |\vec{b}|^2 \]

3. **Rearranging the equation to find \( |\vec{b}|^2 \)**:
\[ |\vec{b}|^2 = 9 - 25 \]
\[ |\vec{b}|^2 = -16 \]

4. **Analyzing the result**:
The square of a magnitude cannot be negative. This indicates that there is no real value for \( |\vec{b}| \) that satisfies the conditions given in the problem. Therefore, we conclude that \( |\vec{b}| \) does not exist in the real number system.

### Final Answer:
There is no valid magnitude for \( |\vec{b}| \) under the given conditions.