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)`

Explanation

To solve the problem step by step, we will use vector addition and the properties of right triangles.

### Step-by-Step Solution:

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

2. **Setting Up the Coordinate System**:
- Place vector \( \vec{A} \) along the x-axis. Thus, we can represent it as \( \vec{A} = A \hat{i} \).
- Since \( \vec{B} \) is added to \( \vec{A} \) and the resultant is along the y-axis, we can represent vector \( \vec{B} \) as \( \vec{B} = 8 \hat{j} \) (assuming it is in the positive y-direction).

3. **Using the Pythagorean Theorem**:
- The resultant vector \( \vec{R} \) can be calculated using the Pythagorean theorem since \( \vec{A} \) and \( \vec{B} \) are perpendicular to each other.
- The magnitude of the resultant vector \( \vec{R} \) is given by:
\[ R = \sqrt{A^2 + B^2} \]
- Substituting \( B = 8 \):
\[ R = \sqrt{A^2 + 8^2} = \sqrt{A^2 + 64} \]

4. **Setting Up the Equation for the Resultant**:
- We know that the magnitude of the resultant vector \( \vec{R} \) is also equal to \( 2A \):
\[ \sqrt{A^2 + 64} = 2A \]

5. **Squaring Both Sides**:
- To eliminate the square root, square both sides:
\[ A^2 + 64 = (2A)^2 \]
- This simplifies to:
\[ A^2 + 64 = 4A^2 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 4A^2 - A^2 - 64 = 0 \]
- This simplifies to:
\[ 3A^2 - 64 = 0 \]

7. **Solving for \( A^2 \)**:
- Rearranging gives:
\[ 3A^2 = 64 \]
- Dividing by 3:
\[ A^2 = \frac{64}{3} \]

8. **Finding the Magnitude of \( A \)**:
- Taking the square root:
\[ A = \sqrt{\frac{64}{3}} = \frac{8}{\sqrt{3}} \]

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