What is the distance between #(-2,1,-7)# and #(11,-5,4)#?

1 Answer
Jun 14, 2018

Answer:

#sqrt326# or about #18.06# (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((11-(-2))^2 + (-5-1)^2 + (4-(-7))^2)#

Now we simplify:
#d = sqrt((13)^2 + (-6)^2 + (11)^2)#

#d = sqrt(169 + 36 + 121)#

#d = sqrt(326)#

If you want to leave it in exact form, you can leave the distance as #sqrt326#. However, if you want the decimal answer, here it is rounded to the nearest hundredth's place:
#d ~~ 18.06#

Hope this helps!