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

2 Answers
Mar 13, 2016

Answer:

#sqrt14#

Explanation:

By using the normal Euclidean metric in #RR^3# we get that

#d[(-4,3,0);(-1,4,2)]=sqrt((-4-(-1))^2+(3-4)^2+(0-(-2))^2)#

#=sqrt(9+1+4)#

#=sqrt14#

Mar 13, 2016

Answer:

Distance between is #sqrt(14)# units

Explanation:

You can construct this scenario using triangles. First you build the xy image (2 space). This image 'if you so prefer, can be viewed as a shadow cast by the actual vector in 3 space. Thus you have two triangles which when combined can be solved using the principle of Pythagoras. Instead of #x^2+y^2# we have differences in #sqrt(x^2+y^2+z^2)#

So for your question we have:

#(x_1,y_1,z_1)-> (-4,3,0)#
#(x_2,y_2,z_2)->(-1,4,-2)#

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

#sqrt((color(white)(.)(-1)-(-4)color(white)(.))^2+(4-3)^2+((-2)-0)^2)#

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