What is the pythagoras-based formula for the distance between two points for a 3D vector?

It follows a very similar format to the pythagoras formula in 2D, in 3D it is: sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

Hi Madeline!

If your two points are (x₁, y₁, z₁) and (x₂, y₂, z₂), then the formula is:

Distance = √[(x₁–x₂)² + (y₁–y₂)² + (z₁–z₂)²]

For example, let's say you want to find the distance between the points (–6, 4, –15) and (2, –5, –3):

Distance = √[(–6–2)² + (4–(–5))² + (–15–(–3))²]

= √[(–8)² + (9)² + (–12)²]

= √[64 + 81 + 144]

= √[289]

= 17

Simon