Vector `vecA,` which is directed along an x axis , is to be added to vector `vecB,` which has a magnitude of `7.0m.` The sum is a third vector that is directed along the y axis, with a magnitude that is `3.0` times that of `vecA.` What is that magnitude of `vecA` ?

Answer

To solve the problem step by step, we will analyze the information given and use vector addition principles.

### Step 1: Understand the vectors
Let vector **A** be directed along the x-axis, and let its magnitude be represented as \( A \). Therefore, we can express vector **A** as:
\[ \vec{A} = A \hat{i} \]
where \( \hat{i} \) is the unit vector in the x-direction.

Vector **B** has a magnitude of \( 7.0 \, m \) and is directed along the y-axis. Thus, we can express vector **B** as:
\[ \vec{B} = 7 \hat{j} \]
where \( \hat{j} \) is the unit vector in the y-direction.

### Step 2: Set up the equation for vector addition
According to the problem, the sum of vectors **A** and **B** results in a vector that is directed along the y-axis. This means that the x-component of the resultant vector must be zero.

The resultant vector **R** can be expressed as:
\[ \vec{R} = \vec{A} + \vec{B} = A \hat{i} + 7 \hat{j} \]
For the resultant vector to be directed along the y-axis, the x-component must equal zero:
\[ A + 0 = 0 \quad \Rightarrow \quad A = -p \]
where \( p \) is the x-component of vector **B**. Since vector **B** is directed along the y-axis, its x-component is zero.

### Step 3: Relate the magnitudes
According to the problem, the magnitude of the resultant vector **R** is \( 3 \) times that of vector **A**. Therefore, we can express this as:
\[ |\vec{R}| = 3A \]
The magnitude of the resultant vector can be calculated using the Pythagorean theorem:
\[ |\vec{R}| = \sqrt{(A)^2 + (7)^2} \]
Substituting the expression for the magnitude of **R**:
\[ \sqrt{A^2 + 7^2} = 3A \]

### Step 4: Solve for A
Now we can square both sides to eliminate the square root:
\[ A^2 + 49 = 9A^2 \]
Rearranging the equation gives:
\[ 9A^2 - A^2 - 49 = 0 \quad \Rightarrow \quad 8A^2 - 49 = 0 \]
\[ 8A^2 = 49 \]
\[ A^2 = \frac{49}{8} \]
Taking the square root of both sides:
\[ A = \sqrt{\frac{49}{8}} = \frac{7}{\sqrt{8}} = \frac{7\sqrt{8}}{8} = \frac{7\sqrt{2}}{4} \]

### Step 5: Final answer
Thus, the magnitude of vector **A** is:
\[ A \approx 2.21 \, m \]