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

2 Answers
Apr 3, 2018

Answer:

Distance b/w the pts.=#sqrt5# units.

Explanation:

let the pts. be A(3,-1,1) & B(2,-3,1)
so, By distance formula
#AB=sqrt(((x_2-x_1)^2)+(y_2-y_1)^2+(z_2-z_1)^2)#
#AB=sqrt[(2-3)^2+(-3+1)^2+(1-1)^2]#
#AB=sqrt[1+4+0]#
#AB=sqrt5# units.

Apr 3, 2018

Answer:

The distance between #(3,-1,1)# and #(2,-3,1)# is #sqrt(5)~~2.236#.

Explanation:

If you have a point #(x_1,y_1,z_1)# and another point #(x_2,y_2,z_2)# and you want to know the distance, you can use the distance formula for a normal pair of #(x,y)# points and add a #z# component. The normal formula is #d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#, so when you add a #z# component, it becomes #d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#. For your points, you would say #sqrt((2-3)^2+((-3)-(-1))^2+(1-1)^2)# which simplifies to #sqrt(5) ~~2.236#