Question

What is the distance between `(3, –1, 1)` and #(2, –3, 1) #?

Answer 1

Distance b/w the pts.=`sqrt5` units.

Explanation:

let the pts. be A(3,-1,1) & B(2,-3,1)
so, By distance formula
`AB=sqrt(((x_2-x_1)^2)+(y_2-y_1)^2+(z_2-z_1)^2)`
`AB=sqrt[(2-3)^2+(-3+1)^2+(1-1)^2]`
`AB=sqrt[1+4+0]`
`AB=sqrt5` units.

Answer 2

The distance between `(3,-1,1)` and `(2,-3,1)` is `sqrt(5)~~2.236`.

Explanation:

If you have a point `(x_1,y_1,z_1)` and another point `(x_2,y_2,z_2)` and you want to know the distance, you can use the distance formula for a normal pair of `(x,y)` points and add a `z` component. The normal formula is `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`, so when you add a `z` component, it becomes `d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)`. For your points, you would say `sqrt((2-3)^2+((-3)-(-1))^2+(1-1)^2)` which simplifies to `sqrt(5) ~~2.236`