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

##### 2 Answers
Apr 12, 2018

$\sqrt{26}$

#### Explanation:

The distance is equal to the magnitude of the vector between the two points which can be expressed as: $| \left(\begin{matrix}4 \\ 1 \\ - 3\end{matrix}\right) - \left(\begin{matrix}0 \\ 4 \\ - 2\end{matrix}\right) |$

$| \left(\begin{matrix}4 - 0 \\ 1 - 4 \\ - 3 - \left(- 2\right)\end{matrix}\right) |$

$| \left(\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right) |$

The magnitude is $\sqrt{{\left(4\right)}^{2} + {\left(- 3\right)}^{2} + {\left(- 1\right)}^{2}}$

$\sqrt{16 + 9 + 1}$ = $\sqrt{26}$

Apr 12, 2018

$A B = \sqrt{26}$

#### 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, $A \left(4 , 1 , - 3\right) \mathmr{and} B \left(0 , 4 , - 2\right)$

$\implies A B = \sqrt{{\left(4 - 0\right)}^{2} + {\left(1 - 4\right)}^{2} + {\left(- 3 + 2\right)}^{2}}$

=>AB=sqrt(16+9+1

$\implies A B = \sqrt{26}$