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

1 Answer
Jul 25, 2018

Answer:

The distance is #3sqrt6# or about #7.348# (rounded to nearest thousandth'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((2 - 4)^2 + (-3-4)^2 + (1-2)^2)#

Now we simplify:
#d = sqrt((-2)^2 + (-7)^2 + (-1)^2)#

#d = sqrt(4 + 49 + 1)#

#d = sqrt(54)#

#d = 3sqrt6#

If you want to leave it in exact form, you can leave the distance as #3sqrt6#. However, if you want the decimal answer, here it is rounded to the nearest thousandth's place:
#d ~~ 7.348#

Hope this helps!