Question

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

Answer 1

`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`

Answer 2

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)`