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

Mar 13, 2016

$\sqrt{14}$

#### Explanation:

By using the normal Euclidean metric in ${\mathbb{R}}^{3}$ we get that

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

$= \sqrt{9 + 1 + 4}$

$= \sqrt{14}$

Mar 13, 2016

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:

$\left({x}_{1} , {y}_{1} , {z}_{1}\right) \to \left(- 4 , 3 , 0\right)$
$\left({x}_{2} , {y}_{2} , {z}_{2}\right) \to \left(- 1 , 4 , - 2\right)$

$\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

$\sqrt{{\left(\textcolor{w h i t e}{.} \left(- 1\right) - \left(- 4\right) \textcolor{w h i t e}{.}\right)}^{2} + {\left(4 - 3\right)}^{2} + {\left(\left(- 2\right) - 0\right)}^{2}}$

$\sqrt{{3}^{2} + {1}^{2} + {\left(- 2\right)}^{2}}$