To find a vector that has the same magnitude as $\mathbf{B}$ and is parallel to $\mathbf{A}$, we can use the following steps:

Step 1: Calculate the magnitude of $\mathbf{B}$

The magnitude of a vector $\mathbf{B} = B_x \hat{i} + B_y \hat{j}$ is given by:

$|\mathbf{B}| = \sqrt{B_x^2 + B_y^2}$

For $\mathbf{B} = 3.0 \hat{i} + 4.0 \hat{j}$, we have:

$|\mathbf{B}| = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0$

Step 2: Find the unit vector of $\mathbf{A}$

The unit vector in the direction of $\mathbf{A}$ is given by:

$\hat{A} = \frac{\mathbf{A}}{|\mathbf{A}|}$

For $\mathbf{A} = 8.1 \hat{i} + 1.0 \hat{j}$, we first calculate its magnitude:

$|\mathbf{A}| = \sqrt{(8.1)^2 + (1.0)^2} = \sqrt{65.61 + 1.0} = \sqrt{66.61} \approx 8.15$

Now, the unit vector $\hat{A}$ is:

$\hat{A} = \frac{8.1 \hat{i} + 1.0 \hat{j}}{8.15} \approx 1.00 \hat{i} + 0.12 \hat{j}$

Step 3: Scale the unit vector to have the same magnitude as $\mathbf{B}$

Now, we scale the unit vector $\hat{A}$ to have the same magnitude as $\mathbf{B}$ by multiplying $\hat{A}$ by $|\mathbf{B}|$:

$\mathbf{C} = |\mathbf{B}| \hat{A} = 5.0 (1.00 \hat{i} + 0.12 \hat{j})$

This gives:

$\mathbf{C} = 5.0 \hat{i} + 0.60 \hat{j}$

So, the vector that has the same magnitude as $\mathbf{B}$ and is parallel to $\mathbf{A}$ is:

$\mathbf{C} = 5.0 \hat{i} + 0.60 \hat{j}$

Final Answer:

The vector that has the same magnitude as $\mathbf{B}$ and is parallel to $\mathbf{A}$ is $\mathbf{C} = 5.0 \hat{i} + 0.60 \hat{j}$.

---

Would you like more details on any part of this process or have any questions?

Here are some related questions for you:

1. How do you calculate the magnitude of a vector in 3 dimensions?
2. How can you find the angle between two vectors?
3. What is the difference between a unit vector and a scalar multiple of a vector?
4. How do you resolve a vector into components?
5. What is the significance of a vector being parallel to another vector?

Tip: Always check the units of your vectors and magnitudes; it's important to keep consistent units throughout the problem!