What is the distance between #(1,-3,-5)# and #(-2,3-,4)#?

1 Answer
Jul 17, 2018

Answer:

The distance is #3sqrt10# or about #9.49# (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((-2-1)^2 + (-3-(-3))^2 + (4-(-5))^2)#

(I wasn't sure if the #3-# meant #3# or #-3#, so I assumed it was #-3#)

Now we simplify:
#d = sqrt((-3)^2 + (0)^2 + (9)^2)#

#d = sqrt(9+0+81)#

#d = sqrt(90)#

#d = 3sqrt10#

If you want to leave it in exact form, you can leave the distance as #3sqrt10#. However, if you want the decimal answer, here it is rounded to the nearest hundredth's place:
#d ~~ 9.49#

Hope this helps!