1. **Problem:** Find the direction of vector $\mathbf{C} = 2\mathbf{A} - \mathbf{B}$ where $\mathbf{A} = 12\mathbf{i} - 16\mathbf{j}$ and $\mathbf{B} = -24\mathbf{i} + 10\mathbf{j}$. 2. **Formula:** The direction angle $\theta$ of a vector $\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}$ is given by $$\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$$ where $\theta$ is measured from the positive $x$-axis. 3. **Calculate $\mathbf{C}$:** $$\mathbf{C} = 2\mathbf{A} - \mathbf{B} = 2(12\mathbf{i} - 16\mathbf{j}) - (-24\mathbf{i} + 10\mathbf{j}) = (24 + 24)\mathbf{i} + (-32 - 10)\mathbf{j} = 48\mathbf{i} - 42\mathbf{j}$$ 4. **Find direction angle:** $$\theta = \tan^{-1}\left(\frac{-42}{48}\right) = \tan^{-1}(-0.875) \approx -41.19^\circ$$ Since $\mathbf{C}$ lies in the fourth quadrant (positive $x$, negative $y$), the direction is $360^\circ - 41.19^\circ = 318.81^\circ$ measured counterclockwise from the positive $x$-axis. --- 5. **Problem:** A camel walks 25 km at 30° south of west, then 30 km north. Find the distance between the two oases. 6. **Approach:** Resolve the first displacement into components: - West component: $25 \cos 30^\circ = 25 \times 0.866 = 21.65$ km (west) - South component: $25 \sin 30^\circ = 25 \times 0.5 = 12.5$ km (south) 7. **Second displacement:** 30 km north. 8. **Net displacement components:** - Horizontal (x): $-21.65$ km (west is negative) - Vertical (y): $-12.5 + 30 = 17.5$ km (north positive) 9. **Distance between oases:** $$d = \sqrt{(-21.65)^2 + (17.5)^2} = \sqrt{469.22 + 306.25} = \sqrt{775.47} \approx 27.85 \text{ km}$$ --- 10. **Problem:** Given $A=5.0$, $B=8.0$, and angle $\alpha=30^\circ$, find the scalar (dot) product. 11. **Formula:** $$\mathbf{A} \cdot \mathbf{B} = AB \cos \alpha$$ 12. **Calculate:** $$5.0 \times 8.0 \times \cos 30^\circ = 40 \times 0.866 = 34.64$$ --- 13. **Problem:** Find the torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ where $\mathbf{r} = \mathbf{i} + \mathbf{j}$ m and $\mathbf{F} = 2\mathbf{i} + 3\mathbf{j}$ N. 14. **Formula:** Torque in 2D is scalar: $$\tau = r_x F_y - r_y F_x$$ 15. **Calculate:** $$\tau = (1)(3) - (1)(2) = 3 - 2 = 1 \text{ N}\cdot\text{m}$$ **Final answers:** 1. Direction of $\mathbf{C}$ is approximately $318.81^\circ$ from positive $x$-axis. 2. Distance between oases is approximately 27.85 km. 3. Scalar product is 34.64. 4. Torque about origin is 1 N·m.