What is the distance between #(–2, 2, 6) # and #(–1, 1, 3) #?

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Answer 1 · 2016-07-31T15:42:10

Distance between #(-2,2,6)# and #(-1,1,3)# is #sqrt11=3.317#

The distance between two points #(x_1,y_1,z_1)# and #(x_2,y_2,z_2)# is given by

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

Hence distance between #(-2,2,6)# and #(-1,1,3)# is

#sqrt(((-1)-(-2))^2+(1-2)^2+(3-6)^2)#

= #sqrt((-1+2)^2+(-1)^2+(-3)^2)#

= #sqrt(1^2+1+9)#

= #sqrt11#

= #3.317#

Answer 2 · 2016-07-31T15:50:52

The distance between to points #(x_1, x_2, x_3)# and #(y_1, y_2, y_3)# in the 3-dimensional space is #sqrt((x_1-y_1)^2+ (x_2-y_2)^2+ (x_3-y_3)^2#

Now simply replacing #(x_1, x_2, x_3)# and #(y_1, y_2, y_3)# by the values given we have:

#sqrt((-2+1)^2+ (2-1)^2+ (6-3)^2)=sqrt((-1)^2+ (1)^2+ (3)^2)=sqrt(11)#

which is the distance