# 6. The 3-dimensional Co-ordinate System

We can expand our 2-dimensional (*x*-*y*) coordinate system into a **3-dimensional coordinate system**, using *x-*, *y-*, and *z*-axes.

The ** x-y plane** is horizontal in our diagram above and shaded green. It can also be described using the equation

*z*= 0, since all points on that plane will have 0 for their

*z*-value.

The ** x-z plane** is vertical and shaded pink above. This plane can be described using the equation `y = 0`.

The ** y-z plane** is also vertical and shaded blue. The

*y*-

*z*plane can be described using the equation `x = 0`.

We normally use the 'right-hand orientation' for the 3 axes, with the positive *x*-axis pointing in the direction of the first finger of our right hand, the positive *y*-axis pointing in the direction of our second finger and the positive *z*-axis pointing up in the direction of our thumb.

Continues below ⇩

### Example - Points in 3-D Space

In 3-dimensional space, the point `(2, 3, 5)` is graphed as follows:

To reach the point `(2, 3, 5)`, we move `2` units along the *x*-axis, then `3` units in the *y*-direction, and then up `5` units in the *z*-direction.

You can explore this example in 3D space using this applet:

## Distance in 3-dimensional Space

To find the distance from one point to another in 3-dimensional space, we just extend Pythagoras' Theorem.

**Distance from the Origin**

The general point *P* (*a*,* b*,* c*) is shown on the 3D graph below. The point *N* is directly below *P* on the *x*-*y* plane.

The distance from `(0, 0, 0)` to the point *P *(*a*,* b*,* c*) is given by:

`"distance"\ OP = sqrt (a^2+ b^2+ c^2)`

**Why?**

The point *N* `(a, b, 0)` is shown on the graph. From Pythagoras' Theorem,

`"distance"\ ON = sqrt (a^2+ b^2)`

and squaring both sides gives:

`(ON)^2=a^2+b^2`

Distance *NP* is simply *c* (this is the distance up the *z*-axis for the point *P*).

Applying Pythagoras' Theorem for the triangle *ONP*, we have:

`"distance"\ OP = sqrt ((ON)^2+ c^2)`

`= sqrt (a^2+ b^2+ c^2)`

### Example 1 - Distance from the Origin to a Point

Find the distance from the origin O to the point *B* `(2, 3, 5)`. This is the example from above.

Answer

`"distance"\ OB= sqrt (2^2+ 3^2+5^2)= sqrt (38)=6.16\ "units"`

This is how it looks on the 3-D graph:

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## Distance Between 2 Points in 3 Dimensions

If we have point A (*x*_{1}, *y*_{1}, *z*_{1}) and another point B (*x*_{2}, *y*_{2}, *z*_{2}) then the distance AB between them is given by the formula:

`"distance"\ AB = ` `sqrt ((x_2-x_1)^2+ (y_2-y_1)^2+ (z_2-z_1)^2)`

This is just an extension of the distance formula (from the origin to a point) that we met above.

### Example 2 - Distance between 2 points

Find the distance between the points P (2, 3, 5) and Q (4, -2, 3).

Answer

The point Q `(4, -2, 3)` is shown on the graph and the distance PQ required is indicated in pink:

Using the formula, we have:

`"distance"\ PQ = sqrt ((x_2-x_1)^2+ (y_2-y_1)^2+ (z_2-z_1)^2)`

`= sqrt ((2-4)^2+(3-(-2))^2+(5-3)^2)`

`=5.74\ "units"`

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