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 1Convert 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 2Parametric 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 3Parametric 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 4Convert 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 5Check 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 6Find 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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