IB Maths AA HL
Topic 3 — Geometry & Trigonometry
Paper 1 & 2
~6 min read
HL only
Magnitude of a Vector & Unit Vectors
The magnitude |v| is the vector’s length — found by Pythagoras in 3D: square the components, add, take the square root. A unit vector has length exactly 1, so dividing any vector by its magnitude gives a unit vector pointing the same way.
📘 What you need to know
- Magnitude formula: |v| = √(v₁² + v₂² + v₃²) — in the formula booklet.
- Magnitude is a non-negative scalar — it’s a length.
- Distance between two points A and B is |AB| = √((x₁−x₂)² + (y₁−y₂)² + (z₁−z₂)²).
- Squaring kills negatives — components’ signs don’t matter for magnitude.
- Unit vector: any vector divided by its magnitude. a|a| has length 1 and points in the direction of a.
- Vector of magnitude k in the direction of a: k × a|a|.
- For a velocity vector, the magnitude is the speed.
Magnitude — Pythagoras in 3D
Magnitude formula
|v| = √(v₁² + v₂² + v₃²) where v = (v₁, v₂, v₃)
The same as the 2D length formula, just with one extra term for the z-component. It’s directly Pythagoras applied twice (once in the xy-plane, once with the z-axis).
Distance between two points
Distance from A to B
|AB| = √((xB−xA)² + (yB−yA)² + (zB−zA)²)
Two-step recipe: find the displacement AB = b − a, then take its magnitude. Same answer as plugging straight into the distance formula.
Unit vectors
Definition
|unit vector| = 1
a vector of length exactly 1
Construction
â = a / |a|
divide any vector by its magnitude → keeps direction, length becomes 1
A unit vector strips out the length info and keeps only the direction. Multiply it back by any scalar k to get a vector of length |k| pointing the same way (or opposite if k is negative).
🧭 Recipe — find a unit vector in the direction of v
- Find |v| using the magnitude formula.
- Divide each component of v by |v|.
- Verify by squaring each new component, adding, and taking √ — should give 1.
- To rescale the unit vector to length k, multiply each component by k.
Worked examples
WE 1Magnitude of a 3D vector
Find the exact magnitude of v = 5i + 2j − 4k.
Step 1: Square each component
5² + 2² + (−4)² = 25 + 4 + 16 = 45
Step 2: Take the square root
|v| = √45 = √(9 × 5) = 3√5
|v| = 3√5
the negative sign on −4 disappears when squared
WE 2Distance between two points
The points A and B have coordinates (2, −3, 1) and (6, 1, −2) respectively. Find the exact distance |AB|.
Step 1: Find AB = b − a
AB = (6−2, 1−(−3), −2−1) = (4, 4, −3)
Step 2: Apply the magnitude formula
|AB|² = 4² + 4² + (−3)² = 16 + 16 + 9 = 41
|AB| = √41
WE 3Unit vector in the direction of v
Find the unit vector in the direction of v = 6i − 3j + 2k.
Step 1: Find the magnitude
|v|² = 6² + (−3)² + 2² = 36 + 9 + 4 = 49
|v| = 7
Step 2: Divide each component by 7
unit vector = 67 i − 37 j + 27 k
check: (6/7)² + (3/7)² + (2/7)² = 49/49 = 1 ✓
WE 4Vector of given magnitude in a direction
Find a vector of magnitude 10 in the direction of u = 3i + 4k.
Step 1: Magnitude of u
|u|² = 3² + 0² + 4² = 25 → |u| = 5
Step 2: Unit vector in direction of u
û = (3/5)i + (4/5)k
Step 3: Multiply by 10
10 û = 10 × (3/5)i + 10 × (4/5)k = 6i + 8k
6i + 8k
check: |6i + 8k| = √(36 + 64) = √100 = 10 ✓
WE 5Find unknown given magnitude
The vector a = (2, k, 4) has magnitude 6. Find the possible values of k.
Step 1: Square the magnitude formula
|a|² = 2² + k² + 4² = 4 + k² + 16 = 20 + k²
Step 2: Set |a| = 6 → |a|² = 36
20 + k² = 36 → k² = 16
Step 3: Solve — both signs work
k = ±4
k² = 16 gives both positive and negative roots — magnitude can’t tell direction
WE 6Speed from a velocity vector
A particle has velocity v = 6i + 2j − 9k m/s. Find its speed.
Speed = magnitude of velocity
Step 1: Square each component
|v|² = 6² + 2² + (−9)² = 36 + 4 + 81 = 121
Step 2: Take √
|v| = √121 = 11
speed = 11 m/s
speed is scalar (magnitude only); velocity is the full vector
💡 Top tips
- Square first, signs disappear. Don’t waste effort on negatives.
- Simplify surds at the end: √45 = 3√5; √50 = 5√2.
- Distance between points = magnitude of the displacement. Same calculation, two names.
- Unit vector check: square the new components and add — should always give 1.
- Vector of length k in direction v: just multiply the unit vector by k.
⚠ Common mistakes
- Forgetting to square root. |v|² ≠ |v| — the formula gives the squared magnitude before the √.
- Adding components instead of squaring. |3i + 4j| = 5, not 7.
- Including signs that get squared anyway. (−4)² = 16, same as 4².
- Forgetting the ± when solving k² = c.
- Calling magnitude a vector — magnitude is always a non-negative scalar (a length).
Next note: The Scalar Product. Multiplying two vectors to get a scalar — using either component-by-component (a₁b₁ + a₂b₂ + a₃b₃) or the angle formula |a||b| cos θ. Used everywhere for finding angles and proving perpendicularity.
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