# What is the distance between (8,3,4) and (1,2,5)?

Mar 11, 2016

$\text{distance=} \sqrt{51}$

#### Explanation:

${P}_{1} = \left(8 , 3 , 4\right) \text{ } {P}_{2} = \left(1 , 2 , 5\right)$
$\Delta x = 1 - 8 = - 7$
$\Delta y = 2 - 3 = - 1$
$\Delta z = 5 - 4 = 1$
$\text{distance=} \sqrt{\Delta {x}^{2} + \Delta {y}^{2} + \Delta {z}^{2}}$
$\text{distance:} \sqrt{{\left(- 7\right)}^{2} + {\left(- 1\right)}^{2} + {1}^{2}}$
$\text{distance=} \sqrt{49 + 1 + 1}$
$\text{distance=} \sqrt{51}$