Shapes with Progressive Dimensionality

The four-dimensional version of the distance formula for coordinates $(x,y,z,q)$ is

$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2 + (q_2-q_1)^2} .$

When restricting values in the interval [0,1], the maximum possible difference for each term in the formula above is 1. This can be achieved (for example) using the coordinate values (0, 0, 0, 0) and (1, 1, 1, 1).

$\sqrt{(1-0)^2 + (1-0)^2 + (1-0)^2 + (1-0)^2} = \sqrt{1 + 1 + 1 +1} = \sqrt{4} = 2 .$

---

(The image below is of a hypercube, although the four-dimensional sides are omitted for clarity.)

Also, if we consider (0, 0, 0, 0) our start point, note the vector $\left<1,1,1,0\right>$ represents the diagonal of the cube (already given as a distance of $\sqrt{3}),$ and $\left<0,0,0,1\right>$ represents the vector connecting the far corner of the cube to the corresponding corner of the hypercube (a distance of 1, since it's simply one side of the hypercube). By the dot product $(1)(0) + (1)(0) + (1)(0) + (0)(1) = 0$ they are perpendicular, so we can apply the Pythagorean Theorem to find the distance between the points (0, 0, 0, 0) and (1, 1, 1, 1):

$\sqrt{\sqrt{3}^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$

$\large d_{\text{max},n,l} = l\sqrt{n}$

$\boxed {\large d_{\text{max},4,1} = \sqrt{4} = 2}$

Where $d_{\text{max}}$ is the Maximum distance between any two points in a $n^{th}$ dimensional space and $l$ is the side length of the $n^{th}$ dimensional space

This is all based on the pattern observed for $1,2$ and $3$ dimensions.