# Distance between two points in space

Given two points in the space $$A=(a_1,a_2,a_3)$$ and $$B=(b_1,b_2,b_3)$$ the distance between them is defined as follows:

the distance between the points $$A$$ and $$B$$ is the module of the vector $$\overrightarrow{AB}$$,

$$\text{d}(A,B)=|\overrightarrow{AB}|= \sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}$$$This distance satisfies the following properties: 1. $$\text{d}(A,B)\geqslant0\$$ and $$\ \text{d}(A,B)=0 \Leftrightarrow \ A=B$$ (Positive) 2. $$\text{d}(A,B)=\text{d}(B,A)$$ (Symmetric) 3. $$\text{d}(A,B)\leqslant \text{d}(A,C)+\text{d}(C,B)$$ (Triangular inequality) From this definition we will determine the distance between any two elements of the space, bearing in mind that the distance between any two elements will always be the smallest one. Calculate the distance between the points $$A = (0, 2, 0)$$ and $$B = (7, 2, -1)$$. We can apply the formula: $$\begin{array}{rl} \text{d}(A,B)=& |\overrightarrow{AB}|=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2 +(b_3-a_3)^2} \\ =&\sqrt{(7-0)^2+(2-2)^2+(-1-0)^2}=\sqrt{50} \end{array}$$$