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NotesMath AA HLTopic 3.13Perpendicular & parallel vectors
Back to Math AA HL Topics
3.13.23 min read

Perpendicular & parallel vectors (Math AA HL)

IB Mathematics: Analysis and Approaches · Unit 3

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Contents

  1. 1Perpendicular ⇔ dot product = 0
  2. 2Parallel ⇔ scalar multiple
A dot product of zero means a right angle: The dot product is a built-in right-angle detector: v·w = 0 ⟺ v and w are perpendicular (also written v ⊥ w).

Why does v·w = 0 mean perpendicular?

1

Start from the angle form

v·w = |v||w| cos θ.

2

The lengths aren't zero

So the only way the product can be zero is cos θ = 0.

3

So the angle is 90°

cos θ = 0 means θ = 90°.

Perpendicular test — set the dot product to zero.

IB-style question — find an unknown for perpendicularity

Vectors are a = (3, k, 2) and b = (1, −4, 5).

Given that a is perpendicular to b, find the value of k.

Step by step

  1. Perpendicular means the dot product is zero.
  2. Write the dot product with k inside.
  3. Simplify the constant terms.
  4. Solve for k.

Final answer

k = 13/4 (then a·b = 0, so a ⊥ b).

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Parallel vectors point the same way — one is a stretch of the other: Two non-zero vectors are parallel when one is a scalar multiple of the other: v = t w for some number t.

What parallel means

  • The components are then proportional — each entry of v is the same multiple t of the matching entry of w.
  • Parallel means θ = 0° or 180°, so cos θ = ±1.
  • Contrast with perpendicular: this is the opposite extreme to the dot-product-zero case.
Parallel test — matching components are in the same ratio t.

IB-style question — find an unknown for parallel

Vectors are u = (4, 6, −2) and v = (2, m, −1).

Given that u is parallel to v, find the value of m.

Step by step

  1. Parallel means u = t·v for some scalar t. Compare the first components.
  2. Check with the third components (should also give t = 2).
  3. Use t = 2 on the middle components.
  4. Solve for m.

Final answer

m = 3 (then u = 2v, so they are parallel).

IB-style question — perpendicular vector of given length

Find a vector q that is perpendicular to p = (3, 4) and has magnitude 10.

Step by step

  1. Swap and negate one component to get a perpendicular direction (check: (3)(−4)+(4)(3)=0).
  2. Its magnitude is √((−4)²+3²) = 5, so scale by 10/5 = 2 to reach length 10.
  3. Check the magnitude.

Final answer

q = (−8, 6) (or its opposite (8, −6)).

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Let and .

Given that
, show that is perpendicular to .
[2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

3.1.1Distance & midpoint (3D)3.1.2Volume & surface area3.1.3Angles in 3D3.1.4Solids in 3D coordinates
View all Math AA HL topics
Improve your exam techniqueCommand terms, paper structure, and mark-scheme tips for Math AA HL
Previous3.13.1Dot product & angle between vectorsNextVector equation of a line3.14.1

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