A point to start from, plus a direction to walk in: Imagine a drone that passes through a known point and then flies in a fixed straight direction. Every position it can reach is:
r = a + λd
a = the position vector of one point you know is on the line (where you 'start').
d = the direction vector — which way the line points (and how far one step of λ moves you).
λ (lambda) = a number you choose. λ = 0 gives the start point a; λ = 1 takes one step along d; λ = 2 takes two steps; negative λ goes backwards.
The same form works in 2D (two components) and 3D (three components) — you just carry an extra row.
Reading off a, d and points on the line: Given r = a + λd, the direction is the vector multiplied by λ, and a is the constant part.
To find an actual position, put a number in for λ and add the components.
Different a's and different-length d's can describe the SAME line — what matters is the direction (any scalar multiple of d points the same way) and that a sits on the line.
IB-style question — read off a position on the line
A delivery drone's path is modelled by
r = (2, 1) + λ(3, −1), with distances in km.
State the point the path passes through at λ = 0, the direction of travel, and find the drone's position when λ = 4.
Step by step
- At λ = 0 the path is at a (the constant part).
- The direction is the vector multiplied by λ.
- Position at λ = 4: add 4 steps of d to a.
Final answer
It passes through (2, 1); it travels in direction (3, −1); at λ = 4 it is at (14, −3), i.e. 14 km east and 3 km south of the origin.
IB-style question — a 3D path
A submarine moves along r = (1, 0, −20) + λ(2, 3, 1), where the third component is depth in metres (negative = below the surface) and λ is in minutes.
Find its position after 5 minutes.
Step by step
- Substitute λ = 5 into each component.
- Add component by component.
Final answer
After 5 minutes the submarine is at (11, 15, −15) — i.e. 15 m below the surface (it has risen from −20 m toward the surface).
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Two points give you everything you need: If you know two points the line passes through, A and B, you can write the line straight away:
Direction = the vector from A to B: d = B − A (subtract the position vectors, finish minus start).
Start point = use A (or B) for a.
So the line is r = A + λ(B − A).
Why 'finish minus start'? B − A is the journey that takes you from A to B, so it points along the line. Any scalar multiple of it (e.g. half it, or double it) is an equally valid direction.
IB-style question — line through two points (2D)
A cable car travels in a straight line from station A(1, 2) to station B(7, 5) (coordinates in km).
Find a vector equation for its path.
Step by step
- Direction = finish minus start: d = B − A.
- Use A as the start point a.
- Write r = a + λd.
Final answer
r = (1, 2) + λ(6, 3). (You could also use direction (2, 1) since it's a scalar multiple of (6, 3), or start from B — all describe the same line.)
IB-style question — line through two points (3D)
A drone flies in a straight line from P(2, −1, 4) to Q(8, 5, 1) (metres).
Find a vector equation for the flight path, and find the position halfway between P and Q.
Step by step
- Direction d = Q − P.
- Vector equation, starting at P.
- Halfway is λ = ½ (a half-step of d from P).
Final answer
r = (2, −1, 4) + λ(6, 6, −3); the midpoint of the flight is (5, 2, 2.5).
IB-style question — where two paths meet
Two drones fly along r₁ = (1, 1) + λ(2, 1) and r₂ = (0, 5) + μ(1, −1) (km).
(a) Find the point where the paths cross.
(b) Find the acute angle between the paths.
Step by step
- (a) Set the position vectors equal, component by component.
- Substitute the first equation into the second and solve for λ.
- Put λ = 1 into r₁ for the crossing point (check: μ = 3 gives the same point in r₂).
- (b) Angle between the directions (2, 1) and (1, −1); take the modulus of the dot product for the acute angle.
Final answer
(a) The paths cross at (3, 2). (b) θ = 71.6°.