What is the distance between #(8, 6, 2) # and #(0, 6, 0) #?

1 Answer
Jan 27, 2016

Answer:

#r=2sqrt(17)#

Explanation:

Let the length of the strait line be r

You can consider the points as a combination of triangles. First you work out the projection of the line on to the xy plain (the adjacent) using Pythagoras. You then work out the related triangle for the z plane again using Pythagoras where r is the hypotenuse (the line). You finish up with a 3 dimensional version of the standard form #r^2=x^2+y^2# except that in the 3d version you have #r^2=x^2+y^2+z^2#

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Given: #(x,y,z)-> (8,6,2) " and " (0,6,0)#

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

#=> r = sqrt((0-8)^2 + (6-6)^2+(0-2)^2)#

#r=sqrt(64+0+4) = sqrt(68) = sqrt(2^2xx17)#

#r=2sqrt(17)#