In our previous posts, we discussed addition and subtraction of vectors, multiple of a vector by a number, and linear combination of vectors. In this post, we introduce the norm and distance of two-dimensional and three-dimensional vectors.

### The norm of a vector

Let \(u = (a,b) \) be a two-dimensional real vector. One can calculate the norm of \(u\), denoted by \(\Vert u \Vert \), by the following formula:

$$ \Vert u \Vert = \sqrt{a^2 + b^2}. $$

For example, if \(u = (3,4) \), then its norm is: $$ \Vert u \Vert = \sqrt{3^2 + 4^2} = 5.$$

Note that if \(u = (a,0) \) or \(u = (0,a) \), then \(\Vert u \Vert = \vert a \vert \). In particular, $$\Vert (0,0) \Vert = 0.$$

**Terminological remark**. Some references sometimes refer to a vector’s norm as its length.

A vector is, by definition, a unit vector if its norm is 1. In the following, we give an example of a unit vector.

**Exercise**. Let \(\theta\) be a real number. Show that the following vector is a unit vector:

$$ u = (\cos \theta , \sin \theta).$$

**Solution**. From elementary trigonometry we know the following trigonometric identity: $$ \cos^2 \theta + \sin^2 \theta = 1.$$

Similarly, if \( u = (a,b,c)\) is a three-dimensional real vector, its norm is computed as follows:

$$ \Vert u \Vert = \sqrt{a^2 + b^2 + c^2}.$$

**Exercise**. Compute the norm of the following three-dimensional vectors:

- \( (3,4,12) \).
- \( (\sqrt{3}/3, -\sqrt{3}/3, \sqrt{3}/3) \)

#### Basic properties of the norm of a vector

The norm of a vector has the following basic properties:

\(\Vert u \Vert \geq 0,\) for any vector \(u\) (non-negativity);

\(\Vert u \Vert = 0\) implies that \( u = \vec{0}\) (positive definiteness);

\(\Vert ru \Vert = \vert r \vert \cdot \Vert u \Vert\) (absolute homogeneity), in particular, we have the following: \(\Vert -u \Vert = \Vert u \Vert \);

\( \Vert u + v \Vert \leq \Vert u \Vert + \Vert v \Vert\) (triangle inequality).

### The distance between two vectors

The distance between two vectors (or two points) is defined to be the norm of their difference. In other words, if \(u\) and \(v\) are two-dimensional (three-dimensional) vectors, the distance between them, denoted by \(d(u,v)\) is, by definition, as follows:

$$d(u,v) = \Vert u – v \Vert.$$

**Exercise**. Compute the distance between the following three-dimensional vectors:

$$ i = (1,0,0), j = (0,1,0)$$

Since $$ i – j = (1,-1,0),$$ we have $$\Vert i – j \Vert = \sqrt{2}. $$

#### Basic properties of the distance between two vectors

Basic properties of the distance between two vectors:

- \(d(u,v) \geq 0 \);
- \(d(u,u) = 0\);
- \(d(u,v) = 0\) implies that \(u = v\);
- \(d(u,v) = d(v,u)\);
- \(d(u,w) \leq d(u,v) + d(v,w)\) (triangle inequality).

**Exercise**. Let \(u\) be any nonzero vector. Prove that \(\displaystyle \frac{1}{\Vert u \Vert} u \) is a unit vector.

**Remark**. Normalization is the name of the process of dividing a nonzero vector \(v\) by its norm. It is clear that normalization keeps its direction but converts the length of a vector into 1. A nonzero vector is normalized if it is divided by its own norm.

**Exercise**. Normalize the following vectors:

- \(u = (3,4)\)
- \(u = (12,4,3)\)

**Exercise**. The reader may try to prove the basic properties of the norm and the distance in two-dimensional real space.

**Remark**. See norm and normalization for lexical discussions.