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# 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. ### 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:

3D Space Interactive 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.

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

This is how it looks on the 3-D graph: ## Distance Between 2 Points in 3 Dimensions

If we have point A (x1, y1, z1) and another point B (x2, y2, z2) 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).

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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