]> 5.4 Representations of a Plane in 3 Dimensions

5.4 Representations of a Plane in 3 Dimensions

We now address the question: what is the relation between the different ways to describe a plane --- by points, one point and a vector, or by an equation?

Suppose the points P 1 , P 2 , P 3 lie in plane Q and they are not all on a line.

Then the vectors P 1 P 2 , and P 1 P 3 have direction in Q and an arbitrary point in Q will have the coordinates of O P 1 + s * P 1 P 2 + t * P 1 P 3 for some pair of values ( s , t ) .

This is called a "parametric" representation of the plane with parameters s and t .

( s and t can be considered the components of the point in the plane in the basis given by P 1 P 2 and P 1 P 3 with origin P 1 .)

You can compute a normal to Q by taking the cross product P 1 P 2 × P 1 P 3 .

We abbreviate by defining

N = P 1 P 2 × P 1 P 3 = O P 2 × O P 3 + O P 3 × O P 1 + O P 1 × O P 2

and so the equation of the plane becomes

N · O P = N · O P 1

where O P = ( x , y , z ) .

You can write this out explicitly as

N x x + N y y + N z z = ( N , P 1 ) = ( N x P 1 x + N y P 1 y + N z P 1 z

It is common, but not necessary to "normalize" N , that is to replace it by n with n = N | N | (remember | N | = ( N · N ) 1 / 2 ) here.

In practice, planes are usually described by a normal vector, like n here, and by a point in it.

We started with three points, and obtained a parametric representation of the plane from them. We then found an equation describing the plane from that representation.

If we can go from the description of Q by this equation back to three points in Q , we will be able to go all the way around the circle and find any representation of Q from any other.

There are an infinite number of points in Q and choosing three of them requires making arbitrary decisions to single out three of them.

If N has all three of its components non-zero we can set each pair of variables to zero and solve for the third one. Then the three points will be

( 0 , 0 , N · O P 1 N z ) , ( 0 , N · O P 1 N y , 0 )

and

( N · O P 1 N x , 0 , 0 )

which are the points at which the plane meets the three axes.

From these points you can go around the circle again, and determine any representation of Q .

In the applet here you can enter three arbitrary points, and it will find and picture the plane, show N , ( N , O P 1 ) , and the parametric representation of points on it. You can do all of these things except making the picture, yourself.

 

 

Exercises:

5.2 Write down the equation for the default plane in this applet, and find the three points in that plane which have two 0 coordinates each.

5.3 Start with three random points and go through this procedure to find N and the points where the plane meets the axes.

5.4 Set up a spreadsheet that does this whenever N has all its components non-zero.

When N z is not zero, we can solve the equation of the plane, N · O P = N · O P 1 for z in terms of x and y , getting

z = N x N z x N y N z y + N · O P 1 N z

The coefficients of x and y here are particularly interesting to us. If you fix y , then our three dimensional space becomes a plane. N x N z then represents the slope of the line that is the intersection of our plane and the plane described by the equation: y = constant . The same statement holds after interchanging x and y in this one.

5.5 Find the two slopes (with y fixed and x fixed) h for the planes you describe in exercises 5.3 and 5.4.