By using the the Pythagorean theorem twice, you can show that $d((0,0,0),(1,2,3))=\sqrt{\left(\sqrt{1^2+2^2}\right)^2+3^2}=\sqrt{1^2+2^2+3^2}$.
In general, if you have two points $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ in $\mathbb{R}^n$, you can use the Pythagorean theorem $n-1$ times to show that the distance between them is
$$\sqrt{\displaystyle\sum_{i=1}^n (x_i -y_i)^2}$$
0
Bumbble Comm
On
It's Pythagorean theorem, just like with 2D space.
By using the the Pythagorean theorem twice, you can show that $d((0,0,0),(1,2,3))=\sqrt{\left(\sqrt{1^2+2^2}\right)^2+3^2}=\sqrt{1^2+2^2+3^2}$.
In general, if you have two points $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ in $\mathbb{R}^n$, you can use the Pythagorean theorem $n-1$ times to show that the distance between them is $$\sqrt{\displaystyle\sum_{i=1}^n (x_i -y_i)^2}$$