Find the "distance"' between the " points "? \(\begin{array}{ll}\text { (a) }(4,-1,2,7) & \text { and }(2,3,1,9) ; \\ \text { (b) }(-1,5,-3,2,4) & \text { and }(2,6,2,7,6)\end{array}\)

The distance formula is crucial for determining the space between two points in any dimension. It's a generalization of the 2D Euclidean distance formula you've probably encountered: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
In higher dimensions, we extend this concept by adding more terms for each additional dimension. For example, the distance \(d\) between two points \(P_1(x_1, y_1, z_1, w_1, \.\.)\) and \(P_2(x_2, y_2, z_2, w_2, \.\.)\) in a four-dimensional space would be:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 + (w_2 - w_1)^2}\]
Let's walk through the application of this formula step-by-step to solidify your understanding.

Higher-dimensional geometry extends the concepts of shapes and distances into more than three dimensions. While it may seem abstract, the mathematical principles remain consistent.
We use coordinates to describe locations in 4D, 5D, etc., just like we do in 3D space using \(x, y, z\)-coordinates. For our problem, we need to calculate distances in a 4-dimensional space for part (a) and a 5-dimensional space for part (b). Here’s how you approach it:
- Identify the coordinates of your points.
- Substitute the coordinates into the distance formula extended into the required number of dimensions.
These steps simplify seemingly complex distances into a straightforward calculation, helping you tackle higher-dimensional problems effectively.

Calculating coordinate differences is the first step in finding the distance between points in any dimension. This involves subtracting corresponding coordinates of the two points, squaring each result, and then summing them all up:
- For part (a), given points \( (4,-1,2,7)\) and \( (2,3,1,9)\), we find the differences:
\[ (4-2) = 2, (-1-3) = -4, (2-1) = 1, (7-9) = -2.\]
Squaring these differences, we get:
\(2^2 = 4, (-4)^2 = 16, 1^2 = 1, (-2)^2 = 4\).
- Summing these: \( 4 + 16 + 1 + 4 = 25\).
- The distance is the square root of this sum: \( \sqrt{25} = 5\).
For part (b), repeat the process for points \( (-1,5,-3,2,4)\) and \( (2,6,2,7,6)\) in 5-dimensional space. This breakdown shows the power of coordinate differences in higher-dimensional spaces.