How to find the distance between points in three dimensions

The distance formula in three dimensions

Given two points $A$ and $B$ in three-dimensional space,

$A(x_1,y_1,z_1)$

$B(x_2,y_2,z_2)$

we can calculate the distance between them using the distance formula.

$D=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

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It doesn’t matter which point is $A$ and which point is $B$. The fact that we square the differences inside the square root means that all of our values will be positive, which means we’ll get a positive value for the distance between the points.

Using the distance formula to calculate the distance between two points in 3D space

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Applying the distance formula to identify the location of points and say which point is closer to a plane

Example

Use the distance formula to

1. Find the distance between $(0,1,3)$ and $(-1,4,5)$.

2. Say which of $(0,1,3)$ and $(-1,4,5)$ lies in the $yz$-plane.

3. Say which of $(0,1,3)$ and $(-1,4,5)$ is closer to the $xy$-plane.

For the first part of the question, we’ll use the distance formula to calculate the distance between the points.

$D=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

$D=\sqrt{(-1-0)^2+(4-1)^2+(5-3)^2}$

$D=\sqrt{1+9+4}$

$D=\sqrt{14}$

The fact that we square the differences inside the square root means that all of our values will be positive, which means we’ll get a positive value for the distance between the points.

For the second part of the question, we need to realize that in order for a point to be in the $yz$-plane, its $x$-coordinate must be $0$. With that in mind, we can say that $(0,1,3)$ lies on the $yz$-plane, and that $(-1,4,5)$ does not lie in the $yz$-plane.

For the third part of the question, we need to realize that the $z$-value of the coordinate point will tell us how far the point is from the $xy$-plane. So if we just take the absolute value of the $z$-coordinate for each of our points, we’ll be able to say which one is closer.

Point $(0,1,3)$ has $|z|=|3|=3$

Point $(-1,4,5)$ has $|z|=|5|=5$

Since the absolute value of $z$ in the point $(0,1,3)$ is less than the absolute value of $z$ in the point $(-1,4,5)$, we can say that $(0,1,3)$ is closer to the $xy$-plane.

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