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

Apr 3, 2018

Distance b/w the pts.=$\sqrt{5}$ units.

Explanation:

let the pts. be A(3,-1,1) & B(2,-3,1)
so, By distance formula
$A B = \sqrt{\left({\left({x}_{2} - {x}_{1}\right)}^{2}\right) + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
$A B = \sqrt{{\left(2 - 3\right)}^{2} + {\left(- 3 + 1\right)}^{2} + {\left(1 - 1\right)}^{2}}$
$A B = \sqrt{1 + 4 + 0}$
$A B = \sqrt{5}$ units.

Apr 3, 2018

The distance between $\left(3 , - 1 , 1\right)$ and $\left(2 , - 3 , 1\right)$ is $\sqrt{5} \approx 2.236$.
If you have a point $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and another point $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ and you want to know the distance, you can use the distance formula for a normal pair of $\left(x , y\right)$ points and add a $z$ component. The normal formula is $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$, so when you add a $z$ component, it becomes $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$. For your points, you would say $\sqrt{{\left(2 - 3\right)}^{2} + {\left(\left(- 3\right) - \left(- 1\right)\right)}^{2} + {\left(1 - 1\right)}^{2}}$ which simplifies to $\sqrt{5} \approx 2.236$