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 $(x, y, z)$ .

Maths; Vectors II; KS5 Year 13; 3D coordinates

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 $k$ :

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