IB Maths AA HL
Topic 3 — Geometry & Trigonometry
Paper 1 & 2
~6 min read
HL only
Parallel Vectors
Two vectors are parallel if one is a scalar multiple of the other — same line of direction, possibly different length, possibly opposite. The condition a = kb (for some non-zero scalar k) shows up everywhere in vector geometry, from finding unknowns to proving three points lie on a line.
📘 What you need to know
- Definition: vectors a and b are parallel ⟺ a = kb for some non-zero scalar k.
- Positive scalar → same direction; negative scalar → opposite direction. Both count as parallel.
- How to test: divide each component of a by the corresponding component of b. Same ratio every time? Parallel.
- Factorise to spot parallels: 9i + 6j − 3k = 3(3i + 2j − k) and −6i − 4j + 2k = −2(3i + 2j − k) → both parallel to (3i + 2j − k).
- If told two vectors are parallel, immediately write a = kb and equate components.
- Three points are collinear if any two of AB, BC, AC are parallel.
The parallel test
Parallel condition
a ∥ b ⟺ a = kb for some scalar k ≠ 0
In components, a = (a₁, a₂, a₃) is parallel to b = (b₁, b₂, b₃) when
a₁b₁ = a₂b₂ = a₃b₃ = k
Same ratio across all three components = parallel. Different ratios anywhere = not parallel.
Same direction or opposite?
Positive k
a = +kb
same direction; just rescaled
Negative k
a = −kb
opposite direction; still parallel
“Parallel” in vector maths includes the anti-parallel case — both a = 2b and a = −2b count as parallel. Direction is along the same line either way.
🧭 Recipe — show two vectors are parallel and find the scalar
- Write both in the same form (column or base vector).
- Divide each component of one by the corresponding component of the other.
- If all the ratios match → they’re parallel; the common ratio is the scalar k.
- If the ratios differ → not parallel.
- For unknowns: set up a = kb, equate components, and solve.
Worked examples
WE 1Show two vectors are parallel + find the scalar
Show that a = 3−12 and b = −9i + 3j − 6k are parallel, and find the scalar k such that b = ka.
Step 1: Write both as columns
a = (3, −1, 2); b = (−9, 3, −6)
Step 2: Divide each component of b by a
−9/3 = −3; 3/−1 = −3; −6/2 = −3
Step 3: All ratios equal → parallel; k = −3
b = −3a, so k = −3
negative k means opposite direction — still parallel
WE 2Find unknown for parallel
Given that a = (4, p, −6) is parallel to b = (−2, 3, 3), find the value of p.
Parallel → a = kb, with same k for every component
Step 1: Find k using known components
x: 4 = k(−2) → k = −2
z-check: −6 = k(3) = (−2)(3) = −6 ✓
Step 2: Apply k to the y-component
p = k(3) = (−2)(3)
p = −6
WE 3Determine if two vectors are parallel
Determine whether a = 2i + 5j − 3k and b = −4i − 10j + 9k are parallel.
Step 1: Compute component ratios b/a
−4/2 = −2; −10/5 = −2; 9/−3 = −3
Step 2: Ratios disagree (−2 ≠ −3)
NOT parallel
need the SAME scalar across every component — one mismatch is enough to fail
WE 4Factorise to spot parallel
Show that c = 12i − 8j + 4k and d = −15i + 10j − 5k are parallel by factorising.
Step 1: Take out a common factor from each
c = 4(3i − 2j + k)
d = −5(3i − 2j + k)
Step 2: Both are scalar multiples of the SAME vector → parallel
Step 3: Find k where d = k·c
d = −5(3i − 2j + k) = (−5/4) × 4(3i − 2j + k) = (−5/4) c
d = −54 c
factorising reveals the shared “direction vector” — useful for spotting parallels quickly
WE 5Find a parameter for parallel
Given that a = ti − 6j + 2tk and b = 4i − 3j + tk are parallel, find the value of t.
Step 1: Set a = kb and equate components
x: t = 4k
y: −6 = −3k → k = 2
z: 2t = tk
Step 2: Use k = 2 in the x-equation
t = 4(2) = 8
Step 3: Check z: 2t = tk → 16 = 8(2) = 16 ✓
t = 8
always check all three components agree on the same k
WE 6Show three points are collinear
The points A(1, 2, −1), B(3, 5, 0), and C(7, 11, 2) are given. Show that A, B, and C are collinear.
Step 1: Compute the displacement vectors AB and BC
AB = B − A = (3−1, 5−2, 0−(−1)) = (2, 3, 1)
BC = C − B = (7−3, 11−5, 2−0) = (4, 6, 2)
Step 2: Check ratios
4/2 = 2; 6/3 = 2; 2/1 = 2
Step 3: BC = 2 AB → AB and BC are parallel and share point B
A, B, C are collinear
collinear = on the same line. Parallel displacements + a shared point seals it.
💡 Top tips
- Always write a = kb as soon as you read “parallel” — it sets up the equations instantly.
- Use the easiest known pair first to find k, then check it works for the third component.
- Factorise long vectors. 12i − 8j + 4k = 4(3i − 2j + k) makes parallels jump out.
- Negative scalars still mean parallel. Don’t reject opposite-direction pairs.
- For collinearity, build two displacement vectors that share a point, then test parallel.
⚠ Common mistakes
- Checking only one or two components. All three must give the same scalar.
- Confusing parallel with equal. Parallel means proportional; equal means identical (k = 1).
- Dropping a sign when dividing. −9/3 = −3, not 3.
- Using k = 0. The zero vector is parallel to everything trivially — questions assume non-zero scalars.
- Forgetting the shared point for collinearity. Parallel vectors alone don’t put points on the same line — you need AB and BC (or similar) sharing an endpoint.
Next note: Adding & Subtracting Vectors. The basic rule (add corresponding components), the geometric meaning (nose-to-tail), and the link between sum vectors and the parallelogram law.
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