1. **State the problem:** We are given position vectors of two ships A and B as functions of time $t$ hours: $$\mathbf{r}_A = (1 + 2t)\mathbf{i} + (3 - 3t)\mathbf{j}$$ $$\mathbf{r}_B = (-2 + 4t)\mathbf{i} + (-4 + t)\mathbf{j}$$ We need to find: (c) The bearing on which ship A is sailing. (d) The value of $t$ when ship B is directly south of ship A. (e) The value of $t$ when ship B is directly south-east of ship A. 2. **Find the velocity vector of ship A:** Velocity is the derivative of position with respect to time $t$. $$\mathbf{v}_A = \frac{d}{dt} \mathbf{r}_A = \frac{d}{dt}[(1 + 2t)\mathbf{i} + (3 - 3t)\mathbf{j}] = 2\mathbf{i} - 3\mathbf{j}$$ 3. **Find the bearing of ship A:** The velocity vector $\mathbf{v}_A = 2\mathbf{i} - 3\mathbf{j}$ means 2 units east and 3 units south. Bearing is measured clockwise from north. Calculate the angle $\theta$ between the velocity vector and the north direction (positive $\mathbf{j}$ axis): $$\tan \theta = \frac{\text{east component}}{\text{north component}} = \frac{2}{-3} = -\frac{2}{3}$$ Since the north component is negative, the vector points south-east quadrant. Calculate $\theta$: $$\theta = \arctan\left|\frac{2}{-3}\right| = \arctan\left(\frac{2}{3}\right) \approx 33.69^\circ$$ Since velocity points south-east, bearing is: $$180^\circ - 33.69^\circ = 146.31^\circ$$ But bearing is clockwise from north, so actually: $$\text{Bearing} = 180^\circ + 33.69^\circ = 213.69^\circ$$ **Final bearing:** $213.7^\circ$ (to 1 decimal place) 4. **Find $t$ when ship B is directly south of ship A:** Ship B is directly south of ship A means they have the same east coordinate (same $x$), and ship B's north coordinate is less than ship A's. Set east components equal: $$1 + 2t = -2 + 4t$$ Simplify: $$1 + 2t = -2 + 4t$$ $$1 + 2t - 4t = -2$$ $$1 - 2t = -2$$ $$-2t = -3$$ $$t = \frac{3}{2} = 1.5$$ 5. **Find $t$ when ship B is directly south-east of ship A:** South-east means ship B is south and east of ship A, and the vector from A to B points exactly $45^\circ$ south of east. Vector from A to B: $$\mathbf{r}_B - \mathbf{r}_A = [(-2 + 4t) - (1 + 2t)]\mathbf{i} + [(-4 + t) - (3 - 3t)]\mathbf{j}$$ Simplify: $$= (-3 + 2t)\mathbf{i} + (-7 + 4t)\mathbf{j}$$ For south-east direction, the vector components satisfy: $$\frac{\text{north component}}{\text{east component}} = \tan(-45^\circ) = -1$$ So: $$\frac{-7 + 4t}{-3 + 2t} = -1$$ Cross multiply: $$-7 + 4t = -1(-3 + 2t) = 3 - 2t$$ Simplify: $$4t + 2t = 3 + 7$$ $$6t = 10$$ $$t = \frac{10}{6} = \frac{5}{3} \approx 1.6667$$ **Summary of answers:** - (c) Bearing of ship A: $213.7^\circ$ - (d) $t = 1.5$ hours - (e) $t = \frac{5}{3}$ hours