The problem involves finding a vector in the direction of $(3, 4)$ with a given magnitude of 2.

To solve this, we follow these steps:

Step 1: Find the unit vector in the direction of $(3, 4)$

The formula for the unit vector $\hat{v}$ in the direction of a vector $(x, y)$ is:

$\hat{v} = \frac{(x, y)}{|v|}$

where $|v|$ is the magnitude (length) of the vector. The magnitude of the vector $(3, 4)$ is calculated as:

$|v| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Thus, the unit vector in the direction of $(3, 4)$ is:

$\hat{v} = \frac{(3, 4)}{5} = \left( \frac{3}{5}, \frac{4}{5} \right)$

Step 2: Scale the unit vector by the desired magnitude (2)

To get the vector in the same direction with a length of 2, we multiply the unit vector by 2:

$\text{desired vector} = 2 \times \left( \frac{3}{5}, \frac{4}{5} \right) = \left( \frac{6}{5}, \frac{8}{5} \right)$

Thus, the vector in the direction of $(3, 4)$ with a length of 2 is:

$\left( \frac{6}{5}, \frac{8}{5} \right)$

This matches the result shown in your image. The simplified answer is already in its most reduced form.

Would you like further explanation, or do you have any questions?

Here are some related questions:

1. How do you compute the unit vector for any given vector?
2. What is the significance of the magnitude of a vector?
3. How do you find the direction of a vector from its components?
4. Can you explain how scaling a vector affects its direction?
5. What is the relationship between unit vectors and normalized vectors?

Tip: When finding vectors in a specific direction, always normalize the original vector to ensure proper direction and magnitude.