What is the distance between # (–2, 1, 3) # and #(–2, 0, 1) #?

1 Answer
Apr 7, 2016

Answer:

#sqrt(5)#

Explanation:

By plotting this out in stages and calculating the projected images on the x,y,z planes you end up with a 3 variable equivalent of the Pythagoras theorem

Let the distance between points be #d#

#=> d= sqrt( (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

#=>d =sqrt( [-2-(-2)]^2+[1-0]^2+[3-1]^2)#

#=>d=sqrt(0+1+4)#

#=>d=+-sqrt(5)#

But the negative side of #sqrt(5)# is not logical for this context so we are only interested in #+sqrt(5)#