Table of Contents

The Distance Between Two Points in ThreeDimensional Space
Recall that if $P(x_1, y_1)$ and $Q(x_2,y_2)$ are points such that $(x_1, y_1,), (x_2, y_2) \in \mathbb{R}^2$ then the distance between $P$ and $Q$ can be computed with the formula $\mathrm{distance} = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2}$. For example, if we wanted to compute the distance between the points $(1, 2)$ and $(1, 4)$, then using the distance formula for $\mathbb{R}^2$ we obtain that:
(1)We also note that the distance between any two points $P, Q \in \mathbb{R}^2$ is zero if and only if $P$ and $Q$ represent the same point.
In $\mathbb{R}^3$, we can compute the distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in a similar manner.
Definition 1: If $P(x, y, z)$ and $Q(x, y, z)$ are two points in $\mathbb{R}^3$, then the Distance between $P$ and $Q$ can be computed with the formula $\mathrm{distance} = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2 + (z_2  z_1)^2}$. 
To show that the distance can be calculated in this manner is not difficult using a geometric argument, however, it is a little messy to illustrate in a twodimensional picture, so we will omit the proof. We will now look at some relatively simply properties of the distance between two points in $\mathbb{R}^3$.
Theorem 1: Let $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ be points in $\mathbb{R}^3$, and let $d_{PQ} = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2 + (z_2  z_1)^2}$ be the distance between $P$ and $Q$. Then $\forall P, Q \in \mathbb{R}^n$: a) $d_{PQ} ≥ 0$. b) $d_{PQ} = 0$ if and only if $(x_1, y_1, z_1) = (x_2, y_2, z_2)$. c) $d_{PQ} = d_{QP}$. 
We will prove a) and b).
 Proof a) We note that $P, Q \in \mathbb{R}^3$ and so the coordinates of $P$ and $Q$ are real numbers. Now we know that for all real numbers $x$ that $\sqrt{x} ≥ 0$. Therefore since the sum $(x_2  x_1)^2 + (y_2  y_1)^2 + (z_2  z_1)^2$ is a real number, then $d_{PQ} ≥ 0$.
 Proof b) $\Leftarrow$ Suppose that $d_{PQ} = 0$. Then we have that:
 Since for any real numbers $x, y \in \mathbb{R}$, $(x  y)^2 ≥ 0$ we have that $(x_2  x_1)^2 ≥ 0$, $(y_2  y_1)^2 ≥ 0$ and $(z_2  z_1)^2 ≥ 0$. Without loss of generality, suppose that $(x_2  x_1)^2 > 0$. Then either $(y_2  y_1)^2 < 0$ or $(z_2  z_1)^2 < 0$, a contradiction, and so we have that $(x_2  x_1)^2 = 0$, $(y_2  y_1)^2 = 0$ and $(z_2  z_1)^2 = 0$, which implies that $x_1 = x_2$, $y_1 = y_2$, and $z_1 = z_2$ and so $(x_1, y_1, z_1) = (x_2, y_2, z_2)$.
 $\Rightarrow$ Suppose that $(x_1, y_1, z_1) = (x_2, y_2, z_2)$. Then computing the distance between these points we have that:
Example 1
Compute the distance between the points $(1, 2, 3), (4, 5, 6) \in \mathbb{R}^3$.
Applying the distance formula directly we have that:
(4)Example 2
Compute the distance between the points $(2, 5, 1), (1, 1, 1) \in \mathbb{R}^3$.
Once again, applying the distance formula directly, we get that:
(5)