Question

What is the distance between `(3, 2, 1)` and #(0, 4, –2) #?

Answer 1

The distance is `sqrt22` or about `4.69` (rounded to nearest hundredth's place)

Explanation:

The formula for the distance for 3-dimensional coordinates is similar or 2-dimensional; it is: `d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2)`

We have the two coordinates, so we can plug in the values for `x`, `y`, and `z`:
`d = sqrt((0-3)^2 + (4-2)^2 + (-2-1)^2)`

Now we simplify:
`d = sqrt((-3)^2 + 2^2 + (-3)^2)`

`d = sqrt(9 + 4 + 9)`

`d = sqrt(22)`

If you want to leave it in exact form, you can leave the distance as `sqrt22`. However, if you want the decimal answer, here it is rounded to the nearest hundredth's place:
`d ~~ 4.69`

Hope this helps!