IB Maths AI HL Vector Properties Paper 1 & 2 Parallel ∥ & perpendicular ⟂ ~8 min read

Components of Vectors

Any vector a can be split into two pieces along any chosen reference direction b: the part parallel to b and the part perpendicular to b. The parallel component has magnitude |a| cos θ — the scalar product divided by |b| gives it directly. The perpendicular component has magnitude |a| sin θ — the vector product magnitude divided by |b| gives it. The two components satisfy Pythagoras: (parallel)² + (perpendicular)² = |a|².

📘 What you need to know

Parallel component: magnitude |a| cos θ = a · b|b| — signed (positive if acute, negative if obtuse).
Perpendicular component: magnitude |a| sin θ = |a × b||b| — always non-negative.
Two equivalent approaches: trigonometric (if you know θ and |a|), or scalar/vector-product (if you know coordinates). Pick whichever fits the given data.
Pythagoras check: (parallel)² + (perpendicular)² = |a|² — useful self-verification of both calculations.
None of these formulas are in the booklet — but they’re built from the scalar and vector product formulas, which are. Sketch a diagram to decide whether to use sin or cos.
Physics applications: resolving a force along a direction (horizontal/vertical or along a slope) is exactly the same procedure as finding the parallel component of a vector.

Decomposing one vector along another

Picture two vectors a and b drawn from the same start, with an angle θ between them. Drop a perpendicular from the tip of a onto the line of b: that splits a into two right-angled pieces. The leg along b has length |a| cos θ — the parallel component. The leg perpendicular to b has length |a| sin θ — the perpendicular component. Since these two legs form a right-angled triangle with a as the hypotenuse, the components satisfy Pythagoras: (parallel)² + (perpendicular)² = |a|².

Dropping a perpendicular from the tip of a to the line of b creates a right-angled triangle. The leg along b is the parallel component (length |a| cos θ); the dashed leg is the perpendicular component (length |a| sin θ). Pythagoras on this triangle gives (parallel)² + (perpendicular)² = |a|².

Component formulas (along reference vector b) parallel: |a| cos θ = a · b|b| · perpendicular: |a| sin θ = |a × b||b| cos for parallel; sin for perpendicular — sketch a diagram to be sure which one applies

Two ways to compute: coordinates vs trigonometry

When the vectors are given by their coordinates, use the scalar-product and vector-product formulas directly — no calculator needed for arccos or sin θ. When you’re given magnitudes and an angle, the trigonometric form (|a| cos θ and |a| sin θ) is faster. The two paths give the same answer because they’re derived from the same geometric formulas for the dot and cross products. For mechanics questions — resolving a force into horizontal and vertical pieces — the trig version is almost always quicker.

Sign of the parallel component: positive when the angle θ < 90° (acute), negative when θ > 90° (obtuse), zero when exactly perpendicular. The perpendicular component is always non-negative because it’s a magnitude.

🧭 Recipe — find a component

Decide which component you need: parallel uses cos, perpendicular uses sin. A quick sketch confirms which leg of the right-angled triangle you’re after.
If you have magnitudes and the angle: parallel = |a| cos θ, perpendicular = |a| sin θ. Done.
If you have coordinates: parallel = (a · b) / |b|; perpendicular = |a × b| / |b|.
For mechanics-style resolution: split the force into horizontal (|F| cos θ) and vertical (|F| sin θ) using the angle the force makes with the horizontal.
Optional verification: check (parallel)² + (perpendicular)² = |a|². If they don’t match, retrace.

Worked examples

WE 1

Parallel component using the scalar product

Find the component of a = (3, 1, 1)^T acting parallel to b = (2, 1, 2)^T.

scalar product a · b = 3·2 + 1·1 + 1·2 = 9 magnitude of b |b| = √(4+1+4) = √9 = 3 parallel = (a · b) / |b| = 9 / 3 parallel component = 3 positive ⇒ a leans toward the same side as b (acute angle between them).

WE 2

Perpendicular component using the vector product

Find the component of a = (3, 0, 0)^T acting perpendicular to b = (1, 2, 2)^T.

vector product a × b comp 1: 0·2 − 0·2 = 0 comp 2: 0·1 − 3·2 = −6 comp 3: 3·2 − 0·1 = 6 a × b = (0, −6, 6)^T magnitude |a × b| = √(0+36+36) = √72 = 6√2 |b| = √(1+4+4) = 3 perpendicular = |a × b| / |b| = 6√2 / 3 = 2√2 perpendicular component = 2√2 ≈ 2.83 cross-check: parallel = (a·b)/|b| = 3/3 = 1; check 1² + (2√2)² = 1 + 8 = 9 = |a|² ✓.

WE 3

Both components and Pythagoras verification

Find the components of a = (3, 4, 0)^T acting (a) parallel to b = (4, 3, 0)^T, (b) perpendicular to b. Verify the Pythagorean relation.

(a) parallel: use a · b a · b = 12 + 12 + 0 = 24 |b| = √(16+9+0) = 5 parallel = 24/5 = 4.8 (b) perpendicular: use a × b a × b = (4·0 − 0·3, 0·4 − 3·0, 3·3 − 4·4) = (0, 0, −7) |a × b| = 7 perpendicular = 7/5 = 1.4 verify Pythagoras (4.8)² + (1.4)² = 23.04 + 1.96 = 25 |a|² = 9 + 16 = 25 ✓ parallel = 4.8; perpendicular = 1.4 2D vectors in the xy-plane — k-component is zero throughout, so a × b is purely in the k direction.

WE 4

Components from magnitude and angle

Two vectors a and b have magnitudes |a| = 8 and |b| = 5, with an angle of 60° between them. Find the components of a acting (a) parallel to b, (b) perpendicular to b.

(a) parallel = |a| cos θ parallel = 8 · cos(60°) = 8 · (1/2) = 4 (b) perpendicular = |a| sin θ perpendicular = 8 · sin(60°) = 8 · (√3/2) = 4√3 parallel = 4; perpendicular = 4√3 ≈ 6.93 verify: 4² + (4√3)² = 16 + 48 = 64 = 8² ✓. Notice that |b| didn’t actually enter the answer — only its direction matters.

WE 5

Mechanics — resolving a force

A force F of magnitude 12 N acts at an angle of 30° above the horizontal. Find the horizontal and vertical components of F.

horizontal = |F| cos θ (along the horizontal) F_x = 12·cos(30°) = 12·(√3/2) = 6√3 N ≈ 10.4 N vertical = |F| sin θ (perpendicular to horizontal) F_y = 12·sin(30°) = 12·(1/2) = 6 N F = 6√3 i + 6j (≈ 10.4 i + 6j) N verify: |F| = √(108 + 36) = √144 = 12 N ✓ — magnitude is preserved.

WE 6

Practical 2D application

A boat moves with velocity v = 5i + 3j m/s. A current flows in the direction w = i + j. Find the component of v in the direction of the current.

column form v = (5, 3, 0)^T; w = (1, 1, 0)^T parallel = (v · w) / |w| v · w = 5·1 + 3·1 + 0 = 8 |w| = √(1+1+0) = √2 parallel = 8/√2 = 8√2/2 = 4√2 component in direction of current = 4√2 m/s ≈ 5.66 m/s rationalise the denominator: 8/√2 × √2/√2 = 8√2/2 = 4√2 — IB standard form is no surds in the denominator.

💡 Top tips

Sketch the angle first: a quick triangle showing a, b, and θ settles whether you need cos (parallel) or sin (perpendicular).
Pick the easier path: if coordinates are given, use the scalar/vector-product formulas directly — no need to compute θ.
Pythagoras as a check: the two components should always satisfy (parallel)² + (perpendicular)² = |a|². Use this whenever you compute both.
Mind the direction of the reference vector in mechanics: “horizontal” and “vertical” simply means decomposing along the x-axis and the y-axis.
Rationalise denominators: 8/√2 should be written as 4√2 in your final answer.

⚠ Common mistakes

Using sin instead of cos (or vice versa) — cos for parallel, sin for perpendicular. A diagram prevents the swap.
Forgetting to divide by |b|: writing parallel = a · b instead of (a · b) / |b|.
Mixing parallel and perpendicular formulas: (a · b) is for parallel; |a × b| is for perpendicular — never the other way around.
Returning a vector instead of a magnitude: the perpendicular component is a single number (length), not the cross-product vector itself.
Sign confusion: parallel component can be negative (obtuse angle); perpendicular is always non-negative (it’s a magnitude).

Next up — Geometric Proof with Vectors. Using the parallel test (scalar multiple), the perpendicular test (zero scalar product), and the equal-length test (equal magnitudes), you can prove that a set of points forms a parallelogram, rectangle, rhombus, or square. The midpoint and collinearity formulas extend these tests to richer geometric claims.

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