# What is the distance between (-2,-1,-7) and (11,5,-3)?

Apr 12, 2018

$\sqrt{221} \approx 14.87 \text{ to 2 dec. places}$

#### Explanation:

$\text{using the 3-dimensional version of the "color(blue)"distance formula}$

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

$\text{let "(x_1,y_1,z_1)=(-2,-1,-7)" and }$

$\left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(11 , 5 , - 3\right)$

$d = \sqrt{{\left(11 + 2\right)}^{2} + {\left(5 + 1\right)}^{2} + {\left(- 3 + 7\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{169 + 36 + 16} = \sqrt{221} \approx 14.87$

Apr 12, 2018

$A B = \sqrt{221}$

#### Explanation:

We know that,

If $A \in {\mathbb{R}}^{3} \mathmr{and} B \in {\mathbb{R}}^{3}$ then the distance between

$A \left({x}_{1} , {y}_{1} , {z}_{1}\right) \mathmr{and} B \left({x}_{2} , {y}_{2} , {z}_{2}\right)$, is

$A B = | \vec{A B} | = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

Where, $\vec{A B} = \left({x}_{2} - {x}_{1} , {y}_{2} - {y}_{1} , {z}_{2} - {z}_{1}\right)$.

We have,

$A \left(- 2 , - 1 , - 7\right) \mathmr{and} B \left(11 , 5 , - 3\right)$

So,

$A B = \sqrt{{\left(11 + 2\right)}^{2} + {\left(5 + 1\right)}^{2} + {\left(- 3 + 7\right)}^{2}}$

$A B = \sqrt{169 + 36 + 16}$

$A B = \sqrt{221}$