# What is the distance between (4,2,2) and (5,-3,-1)?

Feb 5, 2016

$d = \sqrt{35}$

#### Explanation:

Imagine a strong light directly above the line such that the z-axis is vertical and the xy-plane is horizontal. The line would cast a shadow onto the xy-plane (Projected image) and it would in all likelihood form a triangle with the x and y axis.

You could use Pythagoras to determine the length of this projection. You could again use Pythagoras to find the true length but this time the z-axis being as if it is the opposite and the projection being the adjacent.

By going through this process you will find that the final equation boils down to:

Let the distance between the points be d

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

$d = \sqrt{{1}^{2} + {\left(- 5\right)}^{2} + {\left(- 3\right)}^{2}}$

$d = \sqrt{35}$