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 1

Show 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 2

Find 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 3

Determine 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 4

Factorise 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 5

Find 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 6

Show 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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