# What is the distance between (8, 6, 2)  and (0, 6, 0) ?

Jan 27, 2016

$r = 2 \sqrt{17}$

#### Explanation:

Let the length of the strait line be r

You can consider the points as a combination of triangles. First you work out the projection of the line on to the xy plain (the adjacent) using Pythagoras. You then work out the related triangle for the z plane again using Pythagoras where r is the hypotenuse (the line). You finish up with a 3 dimensional version of the standard form ${r}^{2} = {x}^{2} + {y}^{2}$ except that in the 3d version you have ${r}^{2} = {x}^{2} + {y}^{2} + {z}^{2}$

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Given: $\left(x , y , z\right) \to \left(8 , 6 , 2\right) \text{ and } \left(0 , 6 , 0\right)$

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

$\implies r = \sqrt{{\left(0 - 8\right)}^{2} + {\left(6 - 6\right)}^{2} + {\left(0 - 2\right)}^{2}}$

$r = \sqrt{64 + 0 + 4} = \sqrt{68} = \sqrt{{2}^{2} \times 17}$

$r = 2 \sqrt{17}$