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

1 Answer
Jul 25, 2018

Answer:

The distance is #5sqrt2# or about #7.071# (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((1 - (-3))^2 + (2 - 5)^2 + (4 - (-1))^2)#

Now we simplify:
#d = sqrt((4)^2 + (-3)^2 + (5)^2)#

#d = sqrt(16 + 9 + 25)#

#d = sqrt(50)#

#d = 5sqrt2#

If you want to leave it in exact form, you can leave the distance as #5sqrt2#. However, if you want the decimal answer, here it is rounded to the nearest thousandth's place:
#d ~~ 7.071#

Hope this helps!