3D coordinates

​​In a nutshell

A 3D Cartesian coordinate grid uses three axes at right angles to each other: $$x$$, $$y$$ and $$z$$. The coordinates of a point in three dimensions are shown as $$(x, y ,z)$$. You can work out distances on a 3D coordinate grid using Pythagoras' theorem.​

3D Cartesian Coordinates

You are familiar with 2D Cartesian coordinates using the $$x$$​- and $$y$$​-axes. The $$z$$-axis is added to represent the third dimension. You can visualise this as $$x$$- and $$y$$-axes drawn flat, with the $$z$$-axis going through this surface. The axes intersect at right angles. The coordinates of a point on a 3D grid are written as $$(x, y, z)$$.

You can see how point ​$$A$$ with coordinates $$(2, 3, -3)$$ is represented in the graph above, as well as $$B (0, 0, 3)$$.​

Finding distances from the origin

Similar to a 2D coordinate system, you can use Pythagoras' theorem to find distances between points on a 3D coordinate system.

To find the distance $$d$$​ of a point $$P$$ from the origin with coordinates $$(x, y, z)$$, use:

​$$\boxed{d=\sqrt{x^2+y^2+z^2}}$$​​

Example 1

Find the distance from the origin to the point $$P \ (4, -3, 1)$$.

Substitute the values for $$x, y$$ and $$z$$:

$$\begin{aligned}\vert OP\vert&=\sqrt{4^2+(-3)^2+1^2}\\&=\sqrt{16+9+1}\\&=\underline{\sqrt{26}}\end{aligned}$$​​

Finding distances between two points

Using Pythagoras' theorem, the distance $$d$$​ between two points $$P_1 (x_1, y_1,z_1)$$ and $$P_2 (x_2, y_2,z_2)$$ is given by:

​$$\boxed{d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}}$$​​

​

Example 2

Find the distance between the points $$A(4,1,2)$$ and $$B(-1,2,1)$$, giving your answer to $$1 \ d.p.$$​

​$$\begin{aligned}\vert AB\vert &=\sqrt{(4-(-1))^2+(1-2)^2+(2-1)^2}\\&=\sqrt{5^2+(-1)^2+1^2}\\&=\sqrt{25+1+1}\\&=\sqrt{27}\\&=\underline{5.2} \ ( 1\ d.p.)\end{aligned}$$​​

Example 3

The coordinates of $$A$$ and $$B$$ are $$(3, 2, 2)$$ and $$(2,4,k)$$ respectively. Given that the distance $$AB$$ is $$\sqrt6$$​ units, find the possible values of $$k$$.

Using the formula for the distance between two points:

$$\begin{aligned}AB=\sqrt{(3-2)^2+(2-4))^2+(2-k)^2}&=\sqrt6\\\sqrt{1^2+(-2)^2+(2-k)^2}&=\sqrt6\\\sqrt{1+4+(4-4k+k^2)}&=\sqrt6\\1+4+4-4k+k^2&=6\\9-4k+k^2&=6\end{aligned}$$​​

Form a quadratic equation and solve for $$k$$:

$$\begin{aligned}9-4k+k^2&=6\\k^2-4k+3&=0\\(k-3)(k-1)&=0\end{aligned} \\\implies \underline{k=3, k=1}$$​​