Distance Formula in Space

Proof of the Distance Formula for any number of dimensions

Lemma 1. The distance between two points $P_1 (x_1, y_1)$ and $P_2 (x_2, y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Proof of lemma 1: Draw a point $P_3 (x_2, y_1)$. Then since $\angle P_1P_3P_2$ is a right angle, we have $(P_1 P_2)^2 = (P_1 P_3)^2 + (P_3 P_2)^2$. But since $P_1 P_3$ is $|x_2 - x_1|$ and $P_3 P_2$ is $|y_2-y_1|$, we have that $P_1P_2$ must be $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.

Lemma 2. The distance formula between two points $P_1 (x_1, y_1, z_1)$ and $P_2 (x_2, y_2, z_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2 + (z_1-z_2)^2}$.

Proof of lemma 2: Draw a point $P_3 (x_2, y_2, z_1)$. Then by the distance formula for 2 dimensions, the distance $P_1P_3$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Now $P_3P_2$ is parallel to the $z-$axis, so its length is $|z_2-z_1|$. Since the $z-$axis is perpendicular to the $xy-$plane, the triangle formed by the three points and the length of $P_1P_2$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2 + (z_1-z_2)^2}$ by the Pythagorean Theorem.

Now for the main theorem:

Let $P_1$$(a_1, a_2, \dots, a_n)$ and $P_2$$(b_1, b_2, \dots, b_n)$ be two points in $\mathbb{R}^n$. The distance $P_1P_2$ is equal to $\sqrt{\sum_1^n{(a_i-b_i)^2}}$.

Proof: This can be proved inductively. We know it holds in $\mathbb{R}^k$ for $k\le 3$. Now assumes it holds for $k=n-1$. So now we can draw $P_3(b_1, \dots, b_{n-1}, a_n)$, and the distance between $P_3$ and $P_1$ is given by the distance formula for $\mathbb{R}^{n-1}$. The distance from $P_3$ to $P_2$ is simply the difference in their $n$th coordinates, and thus the formula holds in any $\mathbb{R}^{n}$.

Distances

Definition

Pythaorean Theorem

How to compute

Higher Dimensions

Applications

Exercises