IB Maths AI HL Vector Equations of Lines Paper 1 & 2 Parametric form ~7 min read

Equation of a Line in Parametric Form

Same line, different packaging. Take the vector equation r = a + λb and split it into its three components: x = x₀ + λl, y = y₀ + λm, z = z₀ + λn. Each coordinate has its own equation, all sharing the same parameter λ.

📘 What you need to know

Parametric form (in booklet):
· x = x₀ + λl
· y = y₀ + λm
· z = z₀ + λn
What the letters mean:
· (x₀, y₀, z₀) = a known point on the line
· li + mj + nk = direction vector of the line
· λ ∈ ℝ = scalar parameter
Same data as vector form: a = (x₀, y₀, z₀), b = (l, m, n). Just three equations instead of one.
Vector ⇒ parametric: read each component off the vector equation.
Parametric ⇒ vector: stack the point as a and the coefficients of λ as b.
Use parametric when you need a specific coordinate (e.g. “where does the line cross z = 0?”) or when substituting into a Cartesian equation.

From vector form to parametric form

Write r = a + λb as (xyz) = (x₀y₀z₀) + λ(lmn), then equate rows. Each row becomes one parametric equation.

Same line, two presentations. The vector form on the left and the three parametric equations on the right contain exactly the same information — pick whichever is easier for the question.

Parametric form of a line x = x₀ + λl, y = y₀ + λm, z = z₀ + λn (x₀, y₀, z₀) = point on line · (l, m, n) = direction vector

From parametric to vector form

Reverse the process: stack the constants into a and stack the coefficients of λ into b. So x = 5 + 2λ, y = −1 − 3λ, z = 4λ becomes r = (5−10) + λ(2−34).

Why use parametric? Finding where the line crosses x = 0 (or y = 0, z = 0) is one-line easy: set that component to 0, solve for λ, plug back into the other two. Same for intersecting a plane ax + by + cz = d — substitute the parametric equations and solve.

🧭 Recipe — parametric form

Identify point & direction — (x₀, y₀, z₀) on the line, (l, m, n) direction.
Write three equations: x = x₀ + λl, y = y₀ + λm, z = z₀ + λn.
To convert vector ⇒ parametric: split each component of r = a + λb.
To convert parametric ⇒ vector: stack constants as a, coefficients of λ as b.
To find a specific point: substitute a value of λ into all three.
To find where the line crosses a plane (like z = 0): set that variable to the given value, solve for λ, substitute back.

Worked examples

WE 1

Parametric form from point & direction

Write the parametric form of the line through (−2, 1, 0) with direction vector (31−4).

vector form first r = (−210) + λ(31−4) split into 3 equations x = −2 + 3λ y = 1 + λ z = −4λ when y₀ = 0 in z, you just get z = −4λ (no constant term).

WE 2

Vector form from parametric

Convert the parametric equations x = 7 − 2λ, y = 3λ, z = −4 + λ into vector form.

stack constants → a a = (70−4) stack coefficients of λ → b b = (−231) r = (70−4) + λ(−231) no constant in y just means y₀ = 0.

WE 3

Parametric form through two points

Write the parametric form of the line through A(1, 2, −3) and B(4, 0, 1).

direction AB = OB − OA AB = (4−1, 0−2, 1−(−3)) = (3, −2, 4) use A as point on line x = 1 + 3λ y = 2 − 2λ z = −3 + 4λ check: λ = 0 → A(1, 2, −3) ✓ λ = 1 → B(4, 0, 1) ✓

WE 4

Where the line crosses the xy-plane

For the line x = 5 + 2λ, y = −1 + 3λ, z = 4 − 2λ, find the point where the line crosses the plane z = 0.

set z = 0 and solve for λ 4 − 2λ = 0 → λ = 2 substitute λ = 2 into x and y x = 5 + 2(2) = 9 y = −1 + 3(2) = 5 crosses xy-plane at (9, 5, 0) parametric form makes this trivial — one equation, one λ.

WE 5

Find λ from a known coordinate

The line x = 2 + λ, y = −4 + 2λ, z = 6 − λ passes through a point with y-coordinate equal to 10. Find the full coordinates of this point.

set y = 10 and solve for λ −4 + 2λ = 10 → 2λ = 14 → λ = 7 substitute λ = 7 x = 2 + 7 = 9 z = 6 − 7 = −1 point is (9, 10, −1)

WE 6

Intersection of two lines

Determine whether the lines l₁: x = 1 + λ, y = 2 + 2λ, z = λ and l₂: x = 4 + μ, y = 8 − μ, z = 3 + 2μ intersect, and if so find the point.

equate components — use different parameters x: 1 + λ = 4 + μ ⇒ λ − μ = 3 …(1) y: 2 + 2λ = 8 − μ ⇒ 2λ + μ = 6 …(2) add (1) + (2) 3λ = 9 → λ = 3, then μ = 0 check z: 1 · λ vs 3 + 2μ z₁ = 3, z₂ = 3 + 0 = 3 ✓ substitute λ = 3 into l₁ (1+3, 2+6, 3) = (4, 8, 3) lines intersect at (4, 8, 3) always verify with the third equation — if it fails, the lines are skew (don’t meet).

💡 Top tips

Constant in each equation = point coordinate; coefficient of λ = direction component. Read them straight off.
Use parametric form when a question fixes one coordinate (e.g. z = 0) — solve for λ in one equation, plug into the others.
Use different parameters (λ, μ) for different lines when checking intersection — they’re independent variables.
For intersection: solve any two of the three equations for λ and μ, then verify with the third. If the third fails, lines don’t intersect.
Sanity check: λ = 0 should give the original point (x₀, y₀, z₀).

⚠ Common mistakes

Using the same parameter for both lines when checking intersection — use λ for one line and μ for the other.
Forgetting to verify the third equation for intersection — solving 2 out of 3 may give “fake” intersections that don’t actually work.
Mixing up constant and coefficient — in x = 7 − 2λ, the point’s x-coord is 7 and the direction’s x-component is −2 (not the other way round).
Missing constants in components — z = −4λ means z₀ = 0, not “no z equation”.
Confusing parametric form with Cartesian form — parametric uses λ; Cartesian eliminates λ via (x−x₀)/l = (y−y₀)/m = (z−z₀)/n.

Next up — Angle Between Two Lines. Once you have two lines (in vector or parametric form), the angle between them depends only on their direction vectors. Use the scalar product formula cos θ = (b₁ · b₂) / (|b₁| |b₂|), and take the absolute value to get the acute angle.

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