A vector `vec(B)` which has a magnitude 8.0 is added to a vector `vec(A)` which lies along the x-axis. The sum of these two vector is a third vector which lies along the y-axis and has a magnitude that is twice the magnitude of `vec(A)`. Find the magnitude of `vec(A)`

Options

A. `(6)/(sqrt(5))`
B. `(12)/(sqrt(5))`
C. `(16)/(sqrt(5))`
D. `(8)/(sqrt(5))`

Explanation

To solve the problem, we need to find the magnitude of vector \( \vec{A} \) given the conditions about vectors \( \vec{A} \), \( \vec{B} \), and their resultant vector \( \vec{R} \).

### Step-by-Step Solution:

1. **Define the Vectors**:
- Let the magnitude of vector \( \vec{A} \) be \( A \).
- The magnitude of vector \( \vec{B} \) is given as \( B = 8.0 \).
- The resultant vector \( \vec{R} \) lies along the y-axis and has a magnitude that is twice the magnitude of \( \vec{A} \), so \( R = 2A \).

2. **Use the Pythagorean Theorem**:
- Since \( \vec{A} \) lies along the x-axis and \( \vec{B} \) is added to it, the resultant vector \( \vec{R} \) can be found using the Pythagorean theorem:
\[ R^2 = A^2 + B^2 \]
- Substituting \( R = 2A \) and \( B = 8 \):
\[ (2A)^2 = A^2 + 8^2 \]

3. **Expand the Equation**:
- Expanding the left side:
\[ 4A^2 = A^2 + 64 \]

4. **Rearranging the Equation**:
- Move \( A^2 \) to the left side:
\[ 4A^2 - A^2 = 64 \]
- This simplifies to:
\[ 3A^2 = 64 \]

5. **Solve for \( A^2 \)**:
- Divide both sides by 3:
\[ A^2 = \frac{64}{3} \]

6. **Find \( A \)**:
- Taking the square root of both sides:
\[ A = \sqrt{\frac{64}{3}} = \frac{8}{\sqrt{3}} \approx 4.62 \]

### Final Answer:
The magnitude of vector \( \vec{A} \) is \( \frac{8}{\sqrt{3}} \) or approximately \( 4.62 \).