1. Problem 5: Find the direction of vector $C = 2A - B$ where $A = 12i - 16j$ and $B = -24i + 10j$. 2. Use the formula for vector operations: $$C = 2A - B = 2(12i - 16j) - (-24i + 10j)$$ 3. Calculate $2A$: $$2A = 24i - 32j$$ 4. Calculate $-B$: $$-B = 24i - 10j$$ 5. Add $2A$ and $-B$: $$C = (24i - 32j) + (24i - 10j) = (24 + 24)i + (-32 - 10)j = 48i - 42j$$ 6. The direction angle $\theta$ of vector $C$ with respect to the positive $x$-axis is given by: $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-42}{48}\right)$$ 7. Calculate the angle: $$\theta = \tan^{-1}(-0.875) \approx -41.19^\circ$$ 8. Since the angle is negative, it means $41.19^\circ$ below the positive $x$-axis (east). The direction of $C$ is $41.19^\circ$ south of east. 9. Problem 6: A camel walks 25 km in a direction $30^\circ$ south of west, then 30 km north. Find the distance between the two oases. 10. Represent the first displacement vector $\vec{d_1}$: magnitude 25 km, direction $30^\circ$ south of west. 11. Break $\vec{d_1}$ into components: $$d_{1x} = 25 \cos(180^\circ - 30^\circ) = 25 \cos 150^\circ = 25 \times (-0.866) = -21.65$$ $$d_{1y} = 25 \sin 150^\circ = 25 \times 0.5 = 12.5$$ Note: Since direction is south of west, $y$ component should be negative, so correct $d_{1y} = -12.5$ km. 12. The second displacement vector $\vec{d_2}$ is 30 km north, so: $$d_{2x} = 0$$ $$d_{2y} = 30$$ 13. Total displacement vector $\vec{D} = \vec{d_1} + \vec{d_2}$: $$D_x = -21.65 + 0 = -21.65$$ $$D_y = -12.5 + 30 = 17.5$$ 14. Distance between oases is magnitude of $\vec{D}$: $$D = \sqrt{(-21.65)^2 + (17.5)^2} = \sqrt{469.22 + 306.25} = \sqrt{775.47} \approx 27.84$$ km 15. Final answers: - Direction of $C$ is approximately $41.19^\circ$ south of east. - Distance between the two oases is approximately 27.84 km.