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

1 Answer
Anthony X
May 6, 2016

I assume that you know the distance formula (square root of sum of corresponding coordinates squared)
Well, that formula can actually be EXTENDED to the third dimension. (This is a very powerful thing in future mathematics)
What that means is that instead of the known

#sqrt((a-b)^2 + (c-d)^2#

We can extend this to be
#sqrt((a-b)^2 + (c-d)^2 + (e-f)^2#

This problem is beginning to look a lot easier huh?
We can just plug in the corresponding values into the formula

#sqrt((0-0)^2 + (0-6)^2 + (8-0)^2#

#sqrt((0)^2 + (-6)^2 + (8)^2)#

This becomes #sqrt(36+64)#

Which is #sqrt(100)#

This would simplify to #10#

ALTERNATIVELY,
We can see that the x value doesn't change (goes from 0 to 0), so we can really just turn this into a 2 dimensional distance formula, meaning we don't have to extend this and just use

#sqrt((a-b)^2 + (c-d)^2#

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