Vector magnitude and direction

Magnitude

The magnitude of a vector is its length, found with the Pythagorean theorem.

2D: |[x, y]| = √(x² + y²)

3D: |[x, y, z]| = √(x² + y² + z²)

Examples:

• |[3, 4]| = √(9 + 16) = 5
• |[−5, 12]| = 13
• |[2, −1, 2]| = √(4 + 1 + 4) = 3

Direction angle

The direction angle θ is measured counterclockwise from the positive x-axis.

tan θ = y/x

But be careful with quadrants: tan⁻¹ only gives angles between −90° and 90°. Use a reference angle and the signs of the components.

Example: v = [−3, 3]

• Reference angle: tan⁻¹(3/3) = 45°
• Quadrant II → θ = 135°

From magnitude and direction to components

x = |v| cos θ, y = |v| sin θ

Example: |v| = 20, θ = 210°

• x = 20 cos 210° ≈ −17.32
• y = 20 sin 210° = −10
• v ≈ [−17.32, −10]

Bearings

In navigation, direction is given as a bearing (clockwise from north). Convert between bearing β and math angle θ:

θ = 90° − β (add 360° if negative)

Example: Bearing 060° → θ = 30°. Bearing 300° → θ = −210° + 360° = 150°.

Unit vectors

A unit vector in the direction of v:

û = v / |v|

Example: v = [6, 8], |v| = 10 → û = [0.6, 0.8]

Direction in 3D

Direction is described by angles with each axis (direction cosines): cos α = x/|v|, and so on.

Worked examples

Example 1: Magnitude of [7, −24]? 25.

Example 2: Direction angle of [4, −4]? Reference 45°, quadrant IV → 315°.

Example 3: Components of 15 N at 60°? [7.5, 12.99].

Common mistakes

• Using tan⁻¹ without checking the quadrant.
• Negative magnitudes. Magnitude is never negative.
• Mixing bearings and math angles.

Try it yourself

1. |[−8, 6]|
2. |[1, 2, 2]|
3. Direction angle of [−2, −2]?
4. Components of a 50 km/h velocity at 120°.
5. Unit vector in the direction of [5, −12].

Answers: 1) 10 2) 3 3) 225° 4) [−25, 43.3] 5) [5/13, −12/13]

Quick recap

• Magnitude: √(x² + y²) (plus z² in 3D).
• Direction angle: use tan⁻¹ and the quadrant (reference angle).
• Components: |v| cos θ and |v| sin θ; a unit vector is v/|v|.