The Distance Between Two Vectors

# The Distance Between Two Vectors

Sometimes we will want to calculate the distance between two vectors or points. We will derive some special properties of distance in Euclidean n-space thusly. Given some vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$, we denote the distance between those two points in the following manner.

 Definition: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$. Then the Distance between $\vec{u}$ and $\vec{v}$ is $d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$.

We will now look at some properties of the distance between points in $\mathbb{R}^n$.

 Theorem 1 (Symmetry Property of Distance): If $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $d(\vec{u}, \vec{v}) = d(\vec{v}, \vec{u})$.
• Proof: We note that $d(\vec{u}, \vec{v}) = || \vec{u} - \vec{v} || = \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$ and that $d(\vec{v}, \vec{u}) = || \vec{v} - \vec{u} || = \sqrt{(v_1 - u_1)^2 + (v_2 - u_2)^2 ... (v_n - u_n)^2}$. To show these are equal, we must only show that $(u_i - v_i)^2 = (v_i - u_i)^2$ for $1 ≤ i ≤ n$ and $i \in \mathbb{N}$.
• Notice that $(u_i - v_i)^2 = u_i^2 - 2u_iv_i + v_i^2 = v_i^2 - 2u_iv_i + 2u_i^2 = (v_i - u_i)^2$. It therefore follows that $d(\vec{u}, \vec{v}) = d(\vec{v}, \vec{u})$ as the value underneath the square roots is equal. $\blacksquare$
 Theorem 2 (Non-Negativity of Distances): If $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $d(\vec{u}, \vec{v}) ≥ 0$.
• Proof: Since $d(\vec{u}, \vec{v}) = || \vec{u} - \vec{v} || = \sqrt{(u_1 - v_1)^2 + (u_2 - v_2)^2 ... (u_n - v_n)^2}$ and $(u_i - v_i)^2 ≥ 0$ for all $1 ≤ i ≤ n$, $i \in \mathbb{N}$ then clearly $d(\vec{u},\vec{v}) ≥ 0$. $\blacksquare$
 Theorem 3 (The Triangle Inequality of Distances): If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$ then $d(\vec{u}, \vec{v}) \leq d(\vec{u}, \vec{w}) + d(\vec{w}, \vec{v})$.
• Proof: We will begin by operating on the lefthand side:
(1)
\begin{align} d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{w} +\vec{w} - \vec{v} \| \\ d(\vec{u}, \vec{v}) = \| (\vec{u} - \vec{w}) + (\vec{w} - \vec{v}) \| \\ d(\vec{u}, \vec{v}) \leq || (\vec{u} - \vec{w}) || + || (\vec{w} - \vec{v}) \| \\ d(\vec{u}, \vec{v}) \leq d(\vec{u}, \vec{w}) + d(\vec{w}, \vec{v}) \quad \blacksquare \end{align}

## Example 1

Determine the Euclidean distance between $\vec{u} = (2, 3, 4, 2)$ and $\vec{v} = (1, -2, 1, 3)$.

Applying the formula given above we get that:

(2)
\begin{align} d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{(2-1)^2 + (3+2)^2 + (4-1)^2 + (2-3)^2} \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{1 + 25 + 9 + 1} \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = \sqrt{36} \\ d(\vec{u}, \vec{v}) = \| \vec{u} - \vec{v} \| = 6 \end{align}

Therefore $d(\vec{u}, \vec{v}) = 6$.