IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read HL only

Equation of a Line in Parametric Form

Same line as r = a + λb, just split into three scalar equations — one each for x, y, and z. The parameter λ stays the same across all three; the equations are coupled by it.

📘 What you need to know

Parametric form: x = x₀ + λl, y = y₀ + λm, z = z₀ + λn (formula booklet).
(x₀, y₀, z₀) is a known point on the line.
li + mj + nk is a direction vector along the line.
Same λ in all three — it’s a single parameter sliding the point along the line.
Convert from vector form: read off the components of a and b straight into the three equations.
Convert to vector form: stack x₀, y₀, z₀ as a and l, m, n as b.
Point on line: substitute the coordinates into all three equations and check one λ works for all.

The parametric form

Parametric equation of a line x = x₀ + λl , y = y₀ + λm , z = z₀ + λn

Each equation says how one coordinate changes with λ. Plug a value of λ into all three — that gives you a point on the line. Different λ → different point.

Converting between vector and parametric

Vector form

r = a + λb

one compact equation with three components

Parametric form

x = x₀ + λl
y = y₀ + λm
z = z₀ + λn

three scalar equations, same λ

The conversion is mechanical: a = (x₀, y₀, z₀) and b = (l, m, n). No algebra needed — just read the components row by row.

🧭 Recipe — write a line in parametric form

Identify the point: extract (x₀, y₀, z₀) from the question (a point on the line).
Identify the direction: extract (l, m, n) — given as a vector or compute as b − a if two points.
Write three equations: x = x₀ + λl, y = y₀ + λm, z = z₀ + λn.
Check at λ = 0: should give back (x₀, y₀, z₀) — quick sanity check.
If asked, simplify the direction by a common factor (optional, doesn’t change the line).

Worked examples

WE 1

Convert from vector form to parametric form

The line l has vector equation r = (5, −3, 2) + λ(1, 4, −2). Write down the parametric form of l.

Read off components: a = (5, −3, 2), b = (1, 4, −2) Apply x = x₀ + λl, y = y₀ + λm, z = z₀ + λn x = 5 + λ; y = −3 + 4λ; z = 2 − 2λ no calculation needed — just transcribe the components into three equations

WE 2

Parametric form given a point and a direction

Find the parametric form of the equation of the line passing through the point (1, −4, 6) with direction vector 2i − 3j + 5k.

Point: (x₀, y₀, z₀) = (1, −4, 6) Direction: (l, m, n) = (2, −3, 5) x = 1 + 2λ; y = −4 − 3λ; z = 6 + 5λ at λ = 0 this gives (1, −4, 6) ✓

WE 3

Parametric form through two points

Find the parametric form of the equation of the line passing through A(2, 5, −1) and B(8, −1, 3).

Step 1: Direction AB = B − A AB = (8−2, −1−5, 3−(−1)) = (6, −6, 4) Step 2: Simplify direction by 2 (optional) Use (3, −3, 2) Step 3: Use A as the anchor point x = 2 + 3λ; y = 5 − 3λ; z = −1 + 2λ simplifying the direction is fine — same line, just re-scales λ

WE 4

Convert from parametric to vector form

A line has parametric equations x = 4 − λ, y = 2λ, z = −3 + 5λ. Write the equation of the line in vector form.

Read off the constant terms → x₀, y₀, z₀ x₀ = 4, y₀ = 0, z₀ = −3 → a = (4, 0, −3) Read off the coefficients of λ → l, m, n l = −1, m = 2, n = 5 → b = (−1, 2, 5) r = (4, 0, −3) + λ(−1, 2, 5) y = 2λ has no constant — that means y₀ = 0

WE 5

Check if a point lies on a parametric line

Determine whether the point P(10, −5, 10) lies on the line with parametric equations x = 1 + 3λ, y = −2 − λ, z = 4 + 2λ.

Substitute coordinates into each equation and solve for λ x: 10 = 1 + 3λ → λ = 3 y: −5 = −2 − λ → λ = 3 ✓ z: 10 = 4 + 2λ → λ = 3 ✓ P lies on the line (λ = 3) all three give the same λ — point is on the line

WE 6

Find the point on a line for a given parameter value

The line l has parametric equations x = −2 + 4λ, y = 7 − 3λ, z = 1 + λ. Find the coordinates of the point on l when λ = −1.

Substitute λ = −1 into each equation x = −2 + 4(−1) = −6 y = 7 − 3(−1) = 10 z = 1 + (−1) = 0 Point (−6, 10, 0) negative λ just walks the line in the opposite direction from the anchor

💡 Top tips

Use the same λ in all three equations — they’re coupled by the parameter.
Constants give the point; coefficients of λ give the direction.
If y = 2λ, then y₀ = 0 — don’t miss the implicit zero.
Sanity check at λ = 0: you should land on (x₀, y₀, z₀).
Simplify direction by a common factor when possible — cleaner answer, same line.

⚠ Common mistakes

Using a different λ in each equation — they must share one parameter.
Sign errors with negative coefficients — y = 7 − 3λ means the y-direction component is −3, not 3.
Mixing up “constant” with “coefficient”. In x = 1 + 3λ: 1 is the point, 3 is the direction.
Concluding “on the line” after one equation works — must verify all three.
Forgetting y₀ = 0 when an equation is just y = (something)λ with no constant.

Next: Equation of a Line in Cartesian Form. Take the parametric equations, rearrange each to make λ the subject, then set them all equal: x − x₀l = y − y₀m = z − z₀n. The third “form” of the same line.

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