What is the distance between #(-1,4,1)# and #(6,-7,-2)#?

1 Answer
Apr 29, 2018

Answer:

#d = sqrt(179)# or #~~ 13.38#

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((-2-1)^2 + (-7-4)^2 + (6-(-1))^2)#

Now we simplify:
#d = sqrt((-3)^2 + (-11)^2 + (7)^2)#

#d = sqrt(9 + 121 + 49)#

#d = sqrt(179)#

If you want to leave it in exact form, you can leave the distance as #sqrt179#. However, if you want the decimal answer, here it is rounded to the nearest hundredth's place:
#d ~~ 13.38#

Hope this helps!