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

1 Answer
Jul 17, 2018

Answer:

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!