What is the distance between #(-2,-1,-7)# and #(11,5,-3)#?

2 Answers
Apr 12, 2018

Answer:

#sqrt221~~14.87" to 2 dec. places"#

Explanation:

#"using the 3-dimensional version of the "color(blue)"distance formula"#

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

#"let "(x_1,y_1,z_1)=(-2,-1,-7)" and "#

#(x_2,y_2,z_2)=(11,5,-3)#

#d=sqrt((11+2)^2+(5+1)^2+(-3+7)^2)#

#color(white)(d)=sqrt(169+36+16)=sqrt221~~14.87#

Apr 12, 2018

Answer:

#AB=sqrt(221)#

Explanation:

We know that,

If #AinRR^3 and BinRR^3# then the distance between

#A(x_1,y_1,z_1) and B(x_2,y_2,z_2)#, is

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

Where, #vec(AB)=(x_2-x_1,y_2-y_1,z_2-z_1)#.

We have,

#A(-2,-1,-7) andB(11,5,-3)#

So,

#AB=sqrt((11+2)^2+(5+1)^2+(-3+7)^2)#

#AB=sqrt(169+36+16)#

#AB=sqrt(221)#