IB Maths AA HL
Topic 3 — Geometry & Trigonometry
Paper 1 & 2
~6 min read
HL only
Equation of a Line in Vector Form
A line in 3D is described by one point on it plus a direction to walk along — written as r = a + λb. Different values of λ give different points on the line; a anchors it, b tells you which way it goes.
📘 What you need to know
- Vector equation: r = a + λb (in the formula booklet).
- r = position vector of any point on the line.
- a = position vector of one known point on the line.
- b = a direction (displacement) vector along the line.
- λ = a scalar parameter (one number for each point).
- Line through two points A and B: take a as either point, b = AB = bpos − apos.
- Equations are not unique: any point on the line works for a, and any non-zero scalar multiple of b is still a valid direction.
- Point on line check: substitute the point’s coordinates and look for one value of λ that works in all three components.
The vector equation
Vector equation of a line
r = a + λb
Think of a as a “starting point” on the line and b as a “step” in the line’s direction. Choosing λ = 0 gives a; λ = 1 lands you one step further; negative λ walks backwards. Every point on the line corresponds to exactly one value of λ.
Anchor (point)
a
any point you know is on the line — coordinates as a position vector
Direction
b
a vector parallel to the line — usually AB if you know two points
Why it’s not unique: any point on the line can replace a, and b, 2b, −5b all point along the same line. So your answer can look different from the mark scheme but still be correct — check by comparing directions and points.
Line through two points
Through points with position vectors a and p
r = a + λ(p − a) or r = p + λ(p − a)
Use either point as the anchor; the direction is the displacement between them. Both forms describe the same line.
Does a point lie on the line?
To check if point Q with position vector q lies on r = a + λb: substitute q for r, write the three component equations, and try to find a single value of λ that satisfies all three. If one value works for all three components — Q is on the line. If not — it isn’t.
🧭 Recipe — vector equation of a line through two points
- Pick an anchor point: choose either A or B as a.
- Find the direction: b = AB = (other point) − (anchor).
- Write: r = a + λb.
- Simplify the direction vector if it has a common factor — optional but cleaner.
- Sanity check: at λ = 0 you should get the anchor; at λ = 1 you should land on the other point (only if you didn’t simplify).
Worked examples
WE 1Vector equation given a point and a direction
Find a vector equation of the line passing through P(3, −2, 5) with direction vector d = 2i + j − 4k.
Apply r = a + λb directly
a = (3, −2, 5); b = (2, 1, −4)
r = (3, −2, 5) + λ(2, 1, −4)
or equivalently: r = (3i − 2j + 5k) + λ(2i + j − 4k)
WE 2Vector equation through two points
Find a vector equation of the line passing through A(1, 4, −2) and B(5, 6, 2).
Step 1: Direction AB = b − a
AB = (5−1, 6−4, 2−(−2)) = (4, 2, 4)
Step 2: Use A as anchor
r = (1, 4, −2) + λ(4, 2, 4)
you can also write the direction as (2, 1, 2) — half of AB still points the same way
WE 3Show a point lies on a line
Determine whether the point P(7, 1, 0) lies on the line with vector equation r = (1, −2, 3) + λ(2, 1, −1).
Set components equal and solve for λ
x: 7 = 1 + 2λ → λ = 3
y: 1 = −2 + λ → λ = 3 ✓
z: 0 = 3 − λ → λ = 3 ✓
P lies on the line (λ = 3)
all three components give the same λ — that’s the test
WE 4Show a point does NOT lie on a line
Determine whether the point Q(5, 9, 4) lies on the line with vector equation r = (2, 0, −1) + λ(1, 3, 2).
Solve each component for λ
x: 5 = 2 + λ → λ = 3
y: 9 = 0 + 3λ → λ = 3 ✓
z: 4 = −1 + 2λ → λ = 5/2 ✗
Q does NOT lie on the line
the z-component contradicts the others — one mismatch is enough to rule it out
WE 5Find a point on a line for a given parameter value
The line l has vector equation r = (4, −1, 2) + λ(−1, 2, 3). Find the coordinates of the point on l when λ = 2.
Substitute λ = 2 into the equation
r = (4, −1, 2) + 2(−1, 2, 3)
= (4, −1, 2) + (−2, 4, 6)
= (4−2, −1+4, 2+6)
Point (2, 3, 8)
each value of λ gives a different point — that’s how the line is generated
WE 6Find an unknown coordinate so a point lies on a line
The point P(7, k, 1) lies on the line with vector equation r = (1, 2, −3) + λ(3, −1, 2). Find the value of k.
Step 1: Use a known component to find λ
x: 7 = 1 + 3λ → λ = 2
Step 2: Check with z
z: 1 = −3 + 2λ → λ = 2 ✓ (consistent)
Step 3: Use λ = 2 in y-component to find k
k = 2 + (−1)(2) = 0
k = 0
always cross-check λ on a second component before solving for the unknown
💡 Top tips
- Different but correct — your answer’s anchor or scaled direction may differ from the mark scheme. Both can be right.
- Simplify the direction if all components share a factor, e.g. (4, 2, 4) → (2, 1, 2).
- For the point-on-line test, find λ from one component, then check it satisfies the others.
- Two points → direction is just b − a (end minus start).
- Compare to y = mx + c: a is like the intercept (a known point), λb is like mx (slide along the line).
⚠ Common mistakes
- Subtracting in the wrong order for the direction vector — AB = b − a, not a − b.
- Confusing the point and the direction — a is a position vector; b is a direction. Don’t swap them.
- Concluding “point on line” after only checking one or two components — you must verify all three.
- Assuming your equation is wrong just because it looks different from the mark scheme. Check whether your a lies on theirs and your b is a scalar multiple of theirs.
- Forgetting λ is a parameter — the equation describes infinitely many points, one for each value of λ.
Next: Equation of a Line in Parametric Form. Same line, different costume — split r = a + λb into three separate scalar equations x = x0 + λl, y = y0 + λm, z = z0 + λn. Useful for solving intersection problems and converting to Cartesian form.
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