IB Maths AI HLVector Equations of LinesPaper 1 & 2Vector form~8 min read
Equation of a Line in Vector Form
A line in 3D needs two things: a point on it, and a direction. The vector equation r = a + λb packages these together — a is the position vector of a known point, b is a direction vector, and λ is a scalar that slides you along the line. Different λ values give different points on the same line.
📘 What you need to know
Formula (in booklet): r = a + λb · r = position vector of any point on the line
· a = position vector of a known point on the line
· b = direction vector
· λ ∈ ℝ = scalar parameter
Line through two points A & P: direction is a − p (or p − a). Then r = a + λ(a − p) or r = p + λ(a − p).
Infinite valid answers: any point on the line works as a; any non-zero scalar multiple of b works as the direction. Your equation may look different to the mark scheme — still correct.
Is point P on the line? Set p = a + λb and solve each component for λ. All three λ values must agree.
Equivalent test: p − a must be a scalar multiple of b.
Compare withy = mx + c: a plays the role of “a point” (like c), and b plays the role of “direction” (like m).
Writing the equation — r = a + λb
Pick any point on the line, write its position vector as a. Pick any direction along the line, write it as b. The line is the set of all points r = a + λb as λ varies over ℝ. λ = 0 gives the point a; λ = 1 gives a + b; negative λ goes the other way along the line.
The line is fixed; the parameter λ moves you along it. λ = 0 places you at a, λ = 1 moves one full b further, λ = −1 moves one b backwards.
Vector equation of a liner = a + λb ⇔ (xyz) = (a1a2a3) + λ(b1b2b3)a is any known point on the line; b is any non-zero direction vector along the line
Checking whether a point lies on the line
Given r = a + λb and a candidate point p, set p = a + λb and solve each component for λ. If all three give the same value of λ, the point is on the line. If any one disagrees, the point is off the line.
Shortcut: p is on the line ⇔ p − a is a scalar multiple of b. If the components of p − a divided by the components of b all give the same number, that number is your λ.
🧭 Recipe — vector equation of a line
Need a point — pick any point on the line and write its position vector as a.
Need a direction — given two points A, P, use b = a − p (or p − a). Given parametric/cartesian form, read the coefficients of the parameter.
Writer = a + λb.
Test a point P: solve each component pi = ai + λbi for λ — all three must match.
Generate a point at parameter value λ = k: substitute λ = k and compute a + kb.
Worked examples
WE 1
Vector equation from a point and a direction
Write a vector equation of the line through the point (3, −1, 4) with direction vector (20−5).
apply r = a + λb directlya = (3−14), b = (20−5)r = (3−14) + λ(20−5)any scalar multiple of (2, 0, −5) works as direction — e.g. (4, 0, −10) or (−2, 0, 5).
WE 2
Vector equation through two points
Find a vector equation of the line through A with position vector 4i − 5k and B with position vector 3i − 3k.
write OA and OB as columnsOA = (40−5), OB = (30−3)direction = AB = OB − OAAB = (−102)use a = OA, b = ABr = (40−5) + λ(−102)OB + λ(−1, 0, 2) is also a valid answer — same line, different base point.
WE 3
Is the point on the line?
Does the point C(2, 0, −1) lie on the line r = (40−5) + λ(−102)?
set c = a + λb, equate componentsx: 4 − λ = 2 → λ = 2y: 0 + 0λ = 0 ✓ (any λ)z: −5 + 2λ = −1 → λ = 2all components give λ = 2yes, C lies on the line (at λ = 2)when a component is 0 + 0λ on both sides, that equation is satisfied for every λ — ignore it for fixing λ.
WE 4
Point NOT on the line
Does D(7, 2, −3) lie on r = (123) + λ(30−2)?
equate componentsx: 1 + 3λ = 7 → λ = 2y: 2 + 0λ = 2 ✓z: 3 − 2λ = −3 → λ = 3x gives λ = 2 but z gives λ = 3 ✗no, D does NOT lie on the linejust one mismatched λ is enough to rule it out.
WE 5
Find a specific point on the line
For r = (25−1) + λ(1−23): (a) find the point at λ = 3; (b) find the value of λ for which the point lies in the plane z = 0.
Show that r1 = (102) + λ(21−1) and r2 = (520) + μ(−4−22) represent the same line.
directions parallel?(−4, −2, 2) = −2 · (2, 1, −1) ✓(5, 2, 0) on r1? solve (5, 2, 0) = a + λbx: 1 + 2λ = 5 → λ = 2y: 0 + λ = 2 → λ = 2 ✓z: 2 − λ = 0 → λ = 2 ✓parallel + share a point → same liner1 and r2 are the same line ✓two conditions: (1) directions are scalar multiples, (2) one line’s base point lies on the other.
💡 Top tips
Many correct answers — your a and your b may not match the mark scheme but still be right. Two equations describe the same line if they share a point and have parallel directions.
Direction vector is non-zero — b = 0 would give just a single point, not a line.
Use the cleanest direction — divide out common factors. e.g. (4, 2, −6) → (2, 1, −3) describes the same line direction.
Check by substituting λ = 0 — the result should be the point a. Quick sanity check.
When testing a point, equate components and solve for λ. If a component gives 0 = 0, ignore it; use the others to pin λ down.
⚠ Common mistakes
Using a position vector as direction — the direction b is OA − OP, not OA or OP individually.
Forgetting to check all 3 components when testing a point — one component matching is not enough; all three must give the same λ.
Declaring two lines “different” because the equations look different — check direction parallelism AND a shared point first.
Mixing parameter letters — if a problem uses λ for one line and μ for another, keep them separate; never assume λ = μ.
Writing r = a + b instead of r = a + λb — without the parameter you describe a single point, not a line.
Next up — Equation of a Line in Parametric Form. The same equation r = a + λb, split into its three components as x = x0 + λl, y = y0 + λm, z = z0 + λn. Same line, different presentation — you’ll see why parametric form is often easier for solving.
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