What is the distance between #(9, 2, 0) # and #(4, 3, 1) #?

2 Answers
Feb 29, 2016

Answer:

#sqrt((9 - 4)^2 + (2 - 3)^2 + (0 - 1)^2) = sqrt(5^2 + 1^2 + 1^2) = 3sqrt3#

Explanation:

The 2D Pythagorean Theorem states that
ncalculators.com

Now consider a 3D cuboid.

http://demonstrations.wolfram.com/PythagoreanTheorem3D/

Applying the 2D Pythagorean Theorem twice gives

#d^2 = a^2 + z^2 = (x^2 + y^2) + z^2 = x^2 + y^2 + z^2#

Substituting the values #x=5#, #y=1#, #z=1# gives

#d^2 = 5^2 + 1^2 + 1^2 = 27#

#d = sqrt27 = 3sqrt3#

Feb 29, 2016

Answer:

#3sqrt(3)#

Explanation:

The distance between any two points given the rectangular coordinates of the points is:
#color(white)("XX")#the square root of
#color(white)("XXXX")#the sum of
#color(white)("XXXXXX")#the squares of
#color(white)("XXXXXXXX")#the difference between each corresponding pair of coordinates.

In this case we have
#{: ("point A",color(white)("XX"),"(",9,",",color(white)("X")2,",",color(white)("X")0,")"), ("point B",color(white)("XX"),"(",4,",",color(white)("X")3,",",color(white)("X")1,")"), ("difference",color(white)("XX"),"(",5,",",-1,",",-1,")"), ("square of diff",color(white)("XX"),"(",25,",",color(white)("X")1,",",color(white)("X")1,")") :}#

distance #=sqrt(25+1+1) =sqrt(27)=3sqrt(3)#