The problem asks for the direction of the vector difference $\vec{A} - \vec{B}$ and requires the result to be expressed in degrees.

Given:

• Vector $\vec{A}$ has a magnitude of 8.00 m and points along the positive x-axis.
• Vector $\vec{B}$ has a magnitude of 15.0 m and makes a 30.0° angle with the x-axis in the first quadrant.

Approach:

1. Decompose vectors into components:

   • $\vec{A}$ lies on the x-axis, so its components are: $A_x = 8.00 \, \text{m}, \quad A_y = 0 \, \text{m}$
   • For $\vec{B}$, we decompose using trigonometry: $B_x = 15.0 \times \cos(30.0^\circ) = 15.0 \times 0.866 = 12.99 \, \text{m}$ $B_y = 15.0 \times \sin(30.0^\circ) = 15.0 \times 0.5 = 7.5 \, \text{m}$

2. Find the components of the difference $\vec{A} - \vec{B}$: $(A - B)_x = A_x - B_x = 8.00 - 12.99 = -4.99 \, \text{m}$ $(A - B)_y = A_y - B_y = 0 - 7.5 = -7.5 \, \text{m}$

3. Calculate the angle of $\vec{A} - \vec{B}$: The angle $\theta$ is measured from the x-axis and can be found using the inverse tangent function: $\theta = \tan^{-1} \left( \frac{(A - B)_y}{(A - B)_x} \right) = \tan^{-1} \left( \frac{-7.5}{-4.99} \right)$ Since both components are negative, the vector lies in the third quadrant, and the angle will be: $\theta = \tan^{-1}(1.503) \approx 56.7^\circ$ This angle is measured from the negative x-axis, so the actual direction with respect to the positive x-axis is: $\theta = 180^\circ - 56.7^\circ = 123.3^\circ$

Thus, the direction of the vector difference $\vec{A} - \vec{B}$ is 123.3°.

Would you like any more details or further explanations?

Here are some related questions you might find helpful:

1. How do you compute vector magnitudes from components?
2. What is the difference between scalar and vector quantities?
3. How do you add vectors graphically versus using components?
4. How do angles change based on vector quadrant location?
5. How does the inverse tangent function work with vectors?

Tip: Always check the quadrant of a vector before finalizing the angle to ensure accuracy!