Same place AND same time — not just crossing tracks: Two planes can fly over the same point on the map and never crash — as long as they pass through it at different times.
A collision needs both objects at the same position at the same value of t. So set the two position vectors equal and solve EACH coordinate for t:
rA(t) = rB(t).
If one value of t satisfies every coordinate equation, they collide at that time. If the coordinate equations give different t-values (or no solution), the paths may still cross on the map, but the objects are never there together — no collision.
IB-style question — do they collide?
Two drones have positions (in metres, t in seconds)
rA(t) = (1 + 2t, 4 + t) and rB(t) = (7 − t, 1 + 2t).
Determine whether the drones collide.
Step by step
- Collision needs the same position at the same t: set the x-components equal.
- Check this t in the y-components.
- At t = 2 the y-coordinates differ (6 ≠ 5), so the drones are NOT at the same point at the same time.
Final answer
No collision — the only t making the x-coordinates match (t = 2) does not make the y-coordinates match, so they are never at the same place at the same time.
Crossing paths ≠ colliding: To check whether the tracks cross (ignoring time), you'd solve the two line equations with two different parameters. To check a collision, you must use the same t in both — same place AND same moment. Exam questions almost always want the collision check.
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Write the gap as a function of time, then find its minimum: Even when two objects never collide, they have a moment when they are closest. Find it like this:
1. Form the displacement between them, rB(t) − rA(t). 2. Its length is the distance d(t) = √( (Δx)² + (Δy)² ). 3. The closest approach is the minimum of d(t). On a GDC, graph d(t) and read off the minimum (its x-coordinate is the time, its y-coordinate is the least distance).
Tip: d(t) is least exactly when d(t)² is least, so you can minimise the (simpler) squared distance instead.
IB-style question — closest approach to a fixed point
A ship has position r(t) = (−4 + 3t, 2 + 4t) km (t in hours). A lighthouse is fixed at L = (10, 12) km.
Find the time when the ship is closest to the lighthouse, and that least distance.
Step by step
- Displacement from lighthouse to ship.
- Square the distance (easier to minimise).
- Expand and collect.
- Minimum of this parabola at t = −b/(2a) (or graph d(t) on the GDC and read the minimum).
- Least distance: substitute t = 3.28.
Final answer
Closest at t = 3.28 h, at a least distance = 5.2 km. (On a GDC: graph d(t) and read the minimum point ≈ (3.28, 5.2).)
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