The velocity of $B$ as observed by $A$ (velocity of $B$ relative to $A$):
$$\mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A$$
Similarly, relative position: $\mathbf{r}_{B/A}(t) = \mathbf{r}_B(t) - \mathbf{r}_A(t)$.
Example 1: Ship $A$ sails at $10\mathbf{i}+3\mathbf{j}$ km/h. Ship $B$ sails at $4\mathbf{i}+7\mathbf{j}$ km/h.
$\mathbf{v}_{B/A} = (4-10)\mathbf{i}+(7-3)\mathbf{j} = -6\mathbf{i}+4\mathbf{j}$ km/h.
From $A$’s perspective, $B$ is moving at $\sqrt{36+16} = \sqrt{52} \approx 7.21$ km/h in the direction making angle $\arctan(-4/6)$ with the negative $x$-axis.
Method:
1. Form $\mathbf{s}(t) = \mathbf{r}B(t) - \mathbf{r}_A(t)$ (relative position).
2. Minimise $|\mathbf{s}(t)|^2 = \mathbf{s}(t)\cdot\mathbf{s}(t)$.
3. Differentiate and set to zero: $\dfrac{d}{dt}|\mathbf{s}|^2 = 2\mathbf{s}\cdot\dot{\mathbf{s}} = 0$.
4. Note: $\dot{\mathbf{s}} = \mathbf{v}{B/A}$ (constant for constant velocities).
5. So minimum when $\mathbf{s}(t)\cdot\mathbf{v}_{B/A} = 0$ (relative position perpendicular to relative velocity).
Example 2: At $t=0$: $A$ at $(0,0)$ with $\mathbf{v}_A = (3,1)$; $B$ at $(10,5)$ with $\mathbf{v}_B = (1,4)$ (km, km/h).
$\mathbf{r}_A = (3t, t)$; $\mathbf{r}_B = (10+t, 5+4t)$.
$\mathbf{s}(t) = (10-2t, 5+3t)$; $\mathbf{v}_{B/A} = (-2, 3)$.
Closest when $\mathbf{s}\cdot\mathbf{v}_{B/A} = 0$:
$(10-2t)(-2)+(5+3t)(3) = -20+4t+15+9t = -5+13t = 0 \Rightarrow t = 5/13$ h.
$|\mathbf{s}(5/13)|^2 = (10-10/13)^2+(5+15/13)^2 = \left(\frac{120}{13}\right)^2+\left(\frac{80}{13}\right)^2 = \frac{14400+6400}{169} = \frac{20800}{169}$.
Minimum distance $= \dfrac{\sqrt{20800}}{13} = \dfrac{40\sqrt{13}}{13} \approx 11.1$ km.
Example 3: A boat can travel at 20 km/h in still water. A current flows at $5\mathbf{i}$ km/h. To travel due north (direction $\mathbf{j}$), what direction should the boat head?
Let heading angle $\theta$ from north (i.e., $\theta$ toward west). Boat velocity (in water): $20(-\sin\theta\,\mathbf{i}+\cos\theta\,\mathbf{j})$.
Resultant: $(5-20\sin\theta)\mathbf{i}+20\cos\theta\,\mathbf{j}$. For due north: $5-20\sin\theta = 0 \Rightarrow \sin\theta = 1/4 \Rightarrow \theta = \arcsin(0.25) \approx 14.5^\circ$.
Northward speed: $20\cos\theta = 20\sqrt{1-1/16} = 20\sqrt{15/16} = 5\sqrt{15} \approx 19.4$ km/h.
KEY TAKEAWAY: Relative velocity is found by subtracting velocity vectors. Closest approach occurs when the relative position vector is perpendicular to the relative velocity.
EXAM TIP: In navigation problems, always define the positive directions clearly and write both the actual velocity and the required direction as vector conditions before solving.
COMMON MISTAKE: Subtracting positions instead of velocities, or treating relative velocity as a scalar (losing direction information).
