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

1 Answer
Jul 17, 2018

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!