IB Maths AI HL Vector Properties Paper 1 & 2 |v| = √(v₁² + v₂² + v₃²) ~7 min read

Magnitude of a Vector & Unit Vectors

The magnitude |v| of a vector is its length — computed by Pythagoras: square the components, sum, then take the square root. For a position vector it’s the distance from the origin; for a displacement vector AB it’s the distance between A and B. A unit vector is one of magnitude 1, found by dividing any vector by its own magnitude. These two ideas together unlock every “find the distance” and “vector in a given direction” question.

📘 What you need to know

Magnitude formula: |v| = √(v₁² + v₂² + v₃²) — given in the formula booklet (geometry & trigonometry).
Always non-negative: a magnitude is a length, so signs of components don’t matter — squaring removes them.
Distance between points: |AB| = √((x_B−x_A)² + (y_B−y_A)² + (z_B−z_A)²) — first find AB, then its magnitude.
Unit vector: a vector with magnitude 1. The unit vector in the direction of a is â = a/|a|.
Vector of given magnitude in a given direction: take the unit vector in that direction, multiply by the desired magnitude.
Notation: |v| uses vertical bars (sometimes called the modulus). |v|² = v · v — useful shortcut when you’ve already computed a dot product.

Magnitude — Pythagoras in disguise

Drawing a vector and its horizontal and vertical components forms a right-angled triangle: the vector is the hypotenuse, the components are the two legs. So the magnitude is just Pythagoras applied to the components — square each component, sum, take the square root. The same formula extends to 3D by including the third component. Because every term is squared, the signs of the components don’t matter; the magnitude is always a non-negative number.

The vector v = (4, 3)^T sits as the hypotenuse of a right-angled triangle whose legs are its components 4 and 3. Pythagoras gives the magnitude: |v| = √(4² + 3²) = 5. The same recipe with three components handles 3D vectors.

Magnitude & unit vector formulas |v| = √(v₁² + v₂² + v₃²) · â = a|a| distance A→B: first find AB = b − a, then take its magnitude

Unit vectors and rescaling

A unit vector has magnitude 1 — it carries direction information without contributing any length. To find the unit vector in the direction of any non-zero vector a, divide by its own magnitude: â = a/|a|. Once you have the unit vector, you can scale it to any magnitude you want: a vector of magnitude k in the same direction as a is kâ. This is the recipe for any “find a vector of magnitude 6 parallel to…” question.

Pythagorean triples in disguise: vectors like (3, 4, 0), (3, 4, 12), (6, 8, 0), (5, 12, 0), (1, 2, 2) all have integer magnitudes (5, 13, 10, 13, 3). Recognising these saves time — you can write the magnitude directly without computing.

🧭 Recipe — magnitude, distance, unit vector

For a magnitude: square each component, sum, then take the square root.
For a distance between two points: first compute the displacement vector (end minus start), then take its magnitude.
For a unit vector in the direction of a: divide every component of a by |a|.
For a vector of magnitude k in a given direction: find the unit vector in that direction, then multiply each component by k.
For an unknown component: write |v|² = sum-of-squares, set equal to the given magnitude squared, solve.

Worked examples

WE 1

Magnitude of a 3D vector

Find the magnitude of the vector v = (6, −3, 2)^T.

apply the magnitude formula |v|² = 6² + (−3)² + 2² = 36 + 9 + 4 = 49 take the square root |v| = √49 |v| = 7 (−3)² = +9 — sign drops when squared. Clean answer because 6² + 3² + 2² = 49 is a perfect square.

WE 2

Magnitude of a 2D vector (Pythagorean triple)

Find the magnitude of the vector p = 5i − 12j.

only two components in 2D |p|² = 5² + (−12)² = 25 + 144 = 169 |p| = 13 5-12-13 triple — recognise it and skip the square root step.

WE 3

Distance between two 3D points

Find the distance between the points A(1, 2, −3) and B(4, 6, 9).

find the displacement AB = b − a AB = (4−1, 6−2, 9−(−3)) = (3, 4, 12)^T magnitude of AB = distance |AB|² = 9 + 16 + 144 = 169 |AB| = 13 3-4-12-13 quadruple — a 3D Pythagorean answer with integer magnitude.

WE 4

Unit vector in a given direction

Find the unit vector in the direction of a = (4, −3, 0)^T.

find |a| |a|² = 16 + 9 + 0 = 25 |a| = 5 divide each component by |a| â = (4/5, −3/5, 0)^T â = (4/5) i − (3/5) j check: (4/5)² + (−3/5)² = 16/25 + 9/25 = 1 ✓ — unit vector has magnitude 1.

WE 5

Vector of specified magnitude in a direction

Find a vector of magnitude 6 in the same direction as b = 2i − 2j + k.

find |b| |b|² = 4 + 4 + 1 = 9 |b| = 3 required vector = 6 · (b / |b|) = (6/3) · b = 2b multiply each component of b by 2 = 2(2i − 2j + k) = 4i − 4j + 2k = (4, −4, 2)^T check magnitude: √(16+16+4) = √36 = 6 ✓.

WE 6

Find an unknown component from the magnitude

The vector v = (3, t, −4)^T has magnitude √50. Find the possible values of t.

square both sides of |v| = √50 |v|² = 50 set up the equation 3² + t² + (−4)² = 50 9 + t² + 16 = 50 solve for t² t² = 25 t = ±5 t = 5 or t = −5 two solutions — squaring loses sign, so always state both unless the question restricts it.

💡 Top tips

Square first, then add, then square-root — in that order, every time. Skipping the squaring leaves signs intact and gives wrong answers.
Look for Pythagorean triples (3-4-5, 5-12-13, 8-15-17, 2-3-6 in 3D giving 7) — integer magnitudes appear often in IB questions.
For “distance between two points”, always go through the displacement vector first — don’t mix up which is “end” and which is “start”.
Unit vector check: square the components of your answer and add — should give 1. If not, you’ve made an arithmetic slip.
For “magnitude k in direction a“: think “k divided by |a|, then multiply by a” — one factor scales, the other gives direction.

⚠ Common mistakes

Forgetting the square root: writing |v| = 49 instead of |v| = 7 in WE 1 — that’s |v|², not |v|.
Adding before squaring: 5 + (−12) = −7, then 49 ≠ |p|² of 169.
Sign-leakage: leaving a negative inside the square root, e.g. writing √(−49) when (−7)² = +49.
Dividing by the wrong magnitude for a unit vector: dividing a by |b| instead of |a|.
Stating only one value of t from t² = 25 — both t = 5 and t = −5 are valid.

Next up — The Scalar Product. The magnitude formula you just used has a useful link to the dot product: v · v = |v|². That identity, combined with the scalar product’s two formulas, sets up every angle-finding and perpendicularity test in the chapter.

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