To find the direction of the vector $\mathbf{A}_2$, represented by the components $A_{x2} = -6.70 \, \text{m}$ and $A_{y2} = -3.00 \, \text{m}$, we use the following formula:

$\theta = \tan^{-1}\left(\frac{A_{y2}}{A_{x2}}\right)$

Substituting the given values:

$\theta = \tan^{-1}\left(\frac{-3.00}{-6.70}\right)$

$\theta = \tan^{-1}(0.4478)$

Calculating this:

$\theta \approx 24.05^\circ$

Since both $A_{x2}$ and $A_{y2}$ are negative, the vector $\mathbf{A}_2$ lies in the third quadrant. To find the correct direction relative to the positive x-axis (measured counterclockwise), we add $180^\circ$ to the angle calculated in the first quadrant:

$\theta = 24.05^\circ + 180^\circ = 204.05^\circ$

So, the direction of the vector $\mathbf{A}_2$ is approximately 204.05 degrees counterclockwise from the positive x-axis.

Would you like further details or have any questions?

Here are 5 related questions to consider:

1. How would the direction change if $A_{x2}$ were positive and $A_{y2}$ remained negative?
2. What would the direction be if the vector were in the first quadrant?
3. How can you verify the direction using the sine and cosine functions?
4. How does the direction change if the vector's magnitude increases but the components remain proportional?
5. What is the direction if both components are zero?

Tip: When dealing with vectors in different quadrants, always check the signs of the components to determine the correct angle relative to the x-axis.