The magnitude of the x-component of vector `vec(A)` is 3 and the magnitude of vector `vec(A)` is 5. What is the magnitude of the y-component of vector `vec(A)` ?

Options

A. 3
B. 4
C. 5
D. 8

Explanation

To find the magnitude of the y-component of vector \(\vec{A}\), we can use the Pythagorean theorem. The relationship between the components of a vector and its magnitude is given by:

\[ |\vec{A}| = \sqrt{(A_x)^2 + (A_y)^2} \]

Where:
- \(|\vec{A}|\) is the magnitude of the vector,
- \(A_x\) is the x-component of the vector,
- \(A_y\) is the y-component of the vector.

### Step-by-Step Solution:

1. **Identify Given Values**:
- The magnitude of vector \(\vec{A}\) is given as 5.
- The x-component \(A_x\) is given as 3.

2. **Set Up the Equation**:
Using the Pythagorean theorem, we can set up the equation:
\[ 5 = \sqrt{(3)^2 + (A_y)^2} \]

3. **Square Both Sides**:
To eliminate the square root, we square both sides of the equation:
\[ 5^2 = (3)^2 + (A_y)^2 \]
This simplifies to:
\[ 25 = 9 + (A_y)^2 \]

4. **Isolate \(A_y^2\)**:
Rearranging the equation to isolate \((A_y)^2\):
\[ (A_y)^2 = 25 - 9 \]
\[ (A_y)^2 = 16 \]

5. **Take the Square Root**:
Now, take the square root of both sides to find \(A_y\):
\[ A_y = \sqrt{16} \]
\[ A_y = 4 \]

6. **Conclusion**:
The magnitude of the y-component of vector \(\vec{A}\) is 4.

### Final Answer:
The magnitude of the y-component of vector \(\vec{A}\) is 4.