Q1. Find the distance between (1, 2, 3) and (4, 6, 8).
d = √[(4−1)² + (6−2)² + (8−3)²]
d = √[9 + 16 + 25] = √50
d = 5√2 units
Add a third axis and everything from 2D geometry extends naturally. Compute the distance between any two points in space.
Enter two points in space to find the distance between them.
This is exactly the 2D distance formula with one extra (z₂−z₁)² term added under the root.
Q1. Find the distance between (1, 2, 3) and (4, 6, 8).
d = √[(4−1)² + (6−2)² + (8−3)²]
d = √[9 + 16 + 25] = √50
d = 5√2 units
| Quantity | Formula |
|---|---|
| Distance in 3D | d = √[(x₂−x₁)²+(y₂−y₁)²+(z₂−z₁)²] |
| Midpoint in 3D | ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) |
d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²], extending the 2D formula with a z-term.
Eight, formed by the three coordinate planes.
(0, y, 0) — both the x and z coordinates are zero.
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