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 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.
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 = √ (a2 + b2 + c2)
Why?
The point N (a, b, 0) is shown on the graph. From Pythagoras' Theorem,
distance ON = √ (a2 + b2)
and squaring both sides gives:
(ON)2 = a2 + b2
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
= √ ( (ON)2 + c2)
= √(a2 + b2 + c2)
Example - 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 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 = √[(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2]
This is just an extension of the distance formula (from the origin to a point) that we met above.
Example
Find the distance between the points P (2, 3, 5) and Q (4, -2, 3).
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