Find vector magnitudes and unit directions in two and three dimensions
Use the sum of squared components, normalise a non-zero vector and distinguish distance from signed displacement.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.7. Paper 1 coordinate geometry and algebra are assumed. Work in two or three dimensions as stated. The general section theorem is not included; midpoint geometry is required. The cross product, shortest distance between skew lines and their common perpendicular are not required. Scalar products include angles between lines and the foot of a perpendicular from a point to a line, including 3D solids. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Subtract positions to find displacement.
- Use Pythagoras and retain exact square-root values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is (−3)²?
What you will learn
- Calculate magnitudes in 2D and 3D and form a unit vector in a stated direction.
- Use displacement to find point-to-point distance and scale a unit direction to a required length.
Pythagoras gives a non-negative magnitude
For a = (x, y)ᵀ, |a| = √(x² + y²). In three dimensions, |a| = √(x² + y² + z²), using the perpendicular coordinate directions. Thus (2, −3, 6)ᵀ has magnitude √(4 + 9 + 36) = 7. A negative component indicates direction along an axis; it does not subtract from the sum of squared lengths. A magnitude is a scalar, while the original vector remains directed.
The distance between A and B is |b − a|, not generally |b| − |a|. With A = (1, −2, 2) and B = (4, 2, 14), →AB = (3, 4, 12)ᵀ, so AB = 13. The magnitude of →OA is 3, a different distance measured from the origin. Choose the required endpoints before applying the magnitude formula.
Normalise a direction before assigning a length
For non-zero a, the unit vector in its direction is a/|a|. The vector (2, −3, 6)ᵀ therefore gives unit direction (2/7, −3/7, 6/7)ᵀ, whose squared magnitude is 1. The opposite unit direction is its negative. To form a vector of length L ≥ 0 in that direction, multiply the unit vector by L, rather than multiplying the original unnormalised vector by L.
The zero vector has magnitude 0 and no defined direction, so dividing it by its magnitude is impossible. For a non-zero scalar k, |ka| = |k||a|; a negative scalar reverses orientation while preserving a positive magnitude. In a straight constant-velocity model, speed is the magnitude of velocity. Distance along that straight path is speed times elapsed time, while displacement also retains direction. A unit direction is dimensionless.
Pause and explain
Can the zero vector have unit direction 0/|0|?
Worked example
Find the magnitude and unit direction of a = (2, −3, 6)ᵀ, then check the unit magnitude.
Show the worked solution
- Calculate |a| = √(2² + (−3)² + 6²) = 7.
- Divide every component by 7 to obtain (2/7, −3/7, 6/7)ᵀ.
- The sum of squared unit components is (4 + 9 + 36)/49 = 1.
Answer Magnitude 7; unit direction (2/7, −3/7, 6/7)ᵀ.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For a = (3, 4, 0)ᵀ, find its unit direction and a vector of length 12 in the same direction.
Give me a hint
Divide by the magnitude before assigning the new length.
Compare my reasoning
- The magnitude is 5, so the unit direction is (3/5, 4/5, 0)ᵀ.
- Multiply by 12 to get (36/5, 48/5, 0)ᵀ.
- Its magnitude is 12 because the squared components sum to 144; the positive scale retains direction.
Unit direction (3/5, 4/5, 0)ᵀ; length-12 vector (36/5, 48/5, 0)ᵀ.
Look for these in your work
- Normalise all components by the original magnitude.
- Check the new vector's length and orientation.
A = (1, −2, 2) and B = (4, 2, 14). Find →AB, AB, a unit vector from A to B and OA.
Give me a hint
Use the endpoints appropriate to each distance.
Compare my reasoning
- →AB = (3, 4, 12)ᵀ and AB = √(9 + 16 + 144) = 13.
- The unit direction from A to B is (3/13, 4/13, 12/13)ᵀ.
- OA = √(1 + 4 + 4) = 3, distinct from the distance AB.
→AB = (3, 4, 12)ᵀ; AB = 13; unit direction (3/13, 4/13, 12/13)ᵀ; OA = 3.
Look for these in your work
- Subtract positions in the requested direction.
- Keep the origin distance separate from point-to-point distance.
A straight constant-velocity model uses v = (−6, 8) m/s. Starting at (2, 1) m, find the speed, unit direction, time to travel 30 m, displacement and final position.
Give me a hint
Find speed from the magnitude, then use the elapsed time to scale velocity.
Compare my reasoning
- Speed is √(36 + 64) = 10 m/s and unit direction is (−3/5, 4/5)ᵀ.
- Travelling 30 m takes 30/10 = 3 s, giving displacement 3v = (−18, 24) m.
- The final position is (2, 1) + (−18, 24) = (−16, 25) m; the displacement magnitude is 30 m.
Speed 10 m/s; unit direction (−3/5, 4/5)ᵀ; time 3 s; displacement (−18, 24) m; final position (−16, 25) m.
Look for these in your work
- Separate the scalar speed from the directed velocity.
- Check distance, time and final position with consistent units.
Common mistakes
- Square each signed component before adding.
- Find distance from the displacement vector, not a difference of position magnitudes.
- The zero vector cannot be normalised to a unit direction.
What can you explain now?
What is the magnitude of (3, 4, 0)ᵀ?
Compare with the explanation
5.
The squared components sum to 25, so the non-negative magnitude is 5.
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Make it stick
Find a displacement, its magnitude and its unit direction. Explain why signed components do not make a negative magnitude and why normalisation excludes the zero vector.
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