# Question about dimensionality of geometric vectors

I have a seemingly basic question about the transformation of units while normalizing a geometric vector. Let's take for example a vector v having components x and y denoting the distance from the beginning of a coordinate system. Therefore the dimension/unit of both x and y is the length - let's call it D. The length of v is the root of sum of their squares, which means the root of D^2, so simply D again. Given that normalization involves dividing x and y by the length of v, we get D/D which gives us a dimensionless quantity. So the normalized vector is dimensionless or am I making a mistake somewhere?

• Yes the only information it carries is direction, the units are part of the magnitude. – Triatticus Dec 22 '16 at 10:52
• That makes sense, although intuition suggests a normalized vector is simply a vector that is scaled. – krojew Dec 22 '16 at 11:07

## 1 Answer

Yes, your intuition is right. A normalized vector has length $1$, without a unit.

In a sense, it lives in a different vector space than the length-based one, just like velocities, frequencies and various other physical quantities live in different vector spaces. You can multiply any length with a normalized vector to obtain a vector of the given length in the given direction, i.e. to convert from a dimensionless vector to one with a length dimension, but it's indeed best to consider this a separate step.