What is the distance between # (2, 3, 5) # and # (2, 7, 4) #?

Question

1285 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-13T23:31:25

#sqrt(17) ~~ 4.123#

The distance between #(x_1, y_1, z_1)# and #(x_2, y_2, z_2)# is given by the formula:

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

So in our case:

#d = sqrt((2-2)^2+(7-3)^2+(4-5)^2)#

#=sqrt(0+16+1)=sqrt(17) ~~ 4.123#

Distance formula for #3# dimensions

Here's how to derive the distance formula for #3# dimensions from Pythagoras theorem:

Given points #(x_1, y_1, z_1)# and #(x_2, y_2, z_2)#, consider the triangle with vertices:

#(x_1, y_1, z_1)#, #(x_2, y_1, z_1)#, #(x_2, y_2, z_1)#

This is a right angled triangle with legs:

#(x_1, y_1, z_1) (x_2, y_1, z_1)# of length #abs(x_2-x_1)#

#(x_2, y_1, z_1) (x_2, y_2, z_1)# of length #abs(y_2-y_1)#

and hypotenuse:

#(x_1, y_1, z_1) (x_2, y_2, z_1)# of length #sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Next consider the triangle with vertices:

#(x_1, y_1, z_1)#, #(x_2, y_2, z_1)#, #(x_2, y_2, z_2)#

This is a right angled triangle with legs:

#(x_1, y_1, z_1) (x_2, y_2, z_1)# of length #sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#(x_2, y_2, z_1) (x_2, y_2, z_2)# of length #abs(z_2-z_1)#

and hypotenuse:

#(x_1, y_1, z_1) (x_1, y_2, z_2)# of length:

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

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