Unit Vector Calculator

THE UNIT VECTOR
Type a vector and press Normalise it.
The division, the rationalising, and the check
Your vector
Write it however the question does: 3i + 4j, <3, 4>, (3, 4) or just 3, 4. Three components for a vector in space: 2i - 3j + 6k. Fractions and decimals are read and stay exact. Holding deletes one character after another.
A component like 2/√13 is not a finished answer. It is rationalised to 2√13/13, which is the form an exam expects.
Tap a box, then use the keys
A unit vector has length exactly 1 and points the same way as the original. This page proves both before printing anything: the components are squared and added in exact fractions and must come to 1, and multiplying back by the length must give your vector again.
How the unit vector is found

This unit vector calculator normalises any vector to length 1 and gives every component as an exact fraction, with the denominator rationalised.

Type <2, 3> and you get (2√13/13, 3√13/13). Not 2/√13, which is the same number in a form most syllabuses will not accept, and not (0.5547, 0.8321), which is neither.

What Normalising Actually Does

Nothing to the direction. That is the whole point.

A unit vector is the original arrow shrunk or stretched until it is one unit long, pointing exactly where it pointed before. It carries direction and nothing else, which is why it turns up wherever direction matters on its own — resolving forces, lighting a surface in graphics, steering a character in a game.

Textbooks write it with a hat: v̂. Programmers say normalise. Same operation.

One Division, Applied to Everything

Find the length, then divide every component by it.

For 3i + 4j the length is 5, so v̂ = 3/5 i + 4/5 j. Check it: (3/5)² + (4/5)² = 9/25 + 16/25 = 1. Done.

The word every is doing work in that sentence. Divide one component and forget another and you have not shortened the arrow, you have turned it. It is the commonest slip in the topic and it produces an answer that looks plausible and points somewhere else.

Where the Marks Actually Go

3i + 4j is the friendly case because its length is a whole number. Change it slightly and the friendliness disappears.

⟨2, 3⟩ has length √13, so the components come out as 2/√13 and 3/√13. Both correct. Neither acceptable, because a surd is not supposed to be left in a denominator.

Multiply top and bottom by √13 and the root moves upstairs:

2/√13 = 2√13/13

Same value, different appearance, and the second one is what earns the mark. Every component this page prints has already been through that step — you never have to look at a root sitting underneath a fraction.

VectorIts lengthUnit vector, in accepted form
3i + 4j 5 (3/5, 4/5)
⟨2, 3⟩ √13 (2√13/13, 3√13/13)
⟨1, 1⟩ √2 (√2/2, √2/2)
2i − 3j + 6k 7 (2/7, −3/7, 6/7)

Decimals Are Not the Answer, They Are a Reading of It

Well-known calculators hand back (0.5547, 0.8321) for ⟨2, 3⟩ and stop there.

That figure is fine for plotting a picture and useless for the next line of algebra. Square those two decimals and you get 0.99999… — close to 1, not equal to it, and the error grows every time the value is reused. The exact components square to exactly 1, and this page proves that in whole-number fractions before it shows you anything.

You Have Been Using Unit Vectors All Along

Write a vector as 3i + 4j and look at what i and j actually are.

i is the vector ⟨1, 0⟩ — one unit long, pointing along the x-axis. j is ⟨0, 1⟩. Both are unit vectors, and k is the third. So 3i + 4j is not shorthand for a pair of numbers; it is an instruction: go three of these, then four of those. The components were never bare numbers, they were multiples of unit vectors.

That is also why normalising matters in practice rather than only in exams. A force of 20 newtons along some awkward diagonal is written as 20 times the unit vector in that direction, which separates how much from which way and lets you deal with them one at a time. Graphics does the same thing with surface normals, and a game does it with movement: find the direction, normalise it, then multiply by the speed. Skip the normalising and a character moving diagonally travels about 1.41 times faster than one moving straight, which is a bug that has shipped in real games.

The One Vector That Cannot Be Normalised

⟨0, 0⟩ has length zero, and dividing by zero is not something you are allowed to do.

There is also nothing to preserve. An arrow of no length is not pointing anywhere, so there is no direction to keep. Some tools return zeros for this, which is worse than refusing: it looks like an answer. When a zero vector turns up in a physics question it usually means the forces cancelled, and that is the finding, not a failure.

How to Use This Unit Vector Calculator

Write the vector however your question does — 3i + 4j, <3, 4>, (3, 4), or plain 3, 4. Three components give a vector in space. Fractions and decimals both work and both stay exact, so ⟨1/2, 1/3⟩ comes back as (3√13/13, 2√13/13).

The band shows the unit vector, its decimals underneath for when you want them, and the length it was divided by — that figure is worked through in detail on the magnitude and direction page.

Show the working opens the division component by component, the rationalising step written out, and the proof. That proof is not decoration: the components are squared and added as whole-number fractions and must total exactly 1, and multiplying the result back by the length must return your original vector. Both must hold or nothing is printed. The number of checks that passed appears with the working.

Unit Vector Calculator FAQ

What is a unit vector?
A vector of length exactly 1 pointing the same way as the original. It carries the direction and nothing else, which is why it is often called the direction vector and written with a hat, as v̂.
How do you find a unit vector?
Divide every component by the length of the vector. For 3i + 4j the length is 5, so the unit vector is 3/5 i + 4/5 j. Each component must be divided by the same number or the direction changes.
What is the unit vector of (2, 3)?
The length is √13, so the unit vector is (2√13/13, 3√13/13). Written as 2/√13 it is the same number but not in accepted form, because a surd should not be left in a denominator.
Why do you have to rationalise the denominator?
Most syllabuses do not accept a root underneath as a finished answer. Multiplying top and bottom by that root moves it upstairs: 2/√13 becomes 2√13/13, which is the same value written the way a marker expects.
Is normalising a vector the same as finding a unit vector?
Yes. Normalising means scaling a vector to length 1 while keeping its direction, and the result is the unit vector. Programmers usually say normalise, textbooks usually say unit vector.
Can the zero vector be normalised?
No. Its length is zero and dividing by zero is not allowed. There is also nothing to preserve: an arrow of no length does not point anywhere, so a tool that returns zeros here is telling you something false.
How do you check a unit vector is correct?
Square its components and add them. The total must be exactly 1. It takes one line and it catches almost every arithmetic slip, which is why it is worth doing before moving on.
How do you find the unit vector of a 3D vector?
Exactly the same way, with a third component. For 2i − 3j + 6k the length is 7, so the unit vector is (2/7, −3/7, 6/7), and those three squared still add to 1.

Divide by the length, move the root upstairs, then square and add to be sure.