Monday, 9 March 2009

Planes in Space- part One

Take your basic bright kid in alg II or pre-calc, or often in calculus, and ask, "What is the intersection of two lines?"... They say, "A point."... good answer.

"Can you write the equation of a line?" Again they are on target.
"If I give you the equation of two lines on the plane; can you find their point of intersection?" The good ones can, and know they can.

Now ask the same bright kids, "What is the intersection of two planes in space?". They answer correctly again, "A line."
So far everything is great, but now we ask them to write the equation of a plane.... uhhh... gee..... and at this point, when asked about one of the fundamental structures of plane geometry, their analytic geometry skills are exhausted. Still, a very few may actually be able to produce x+y+z=1 or some other for the equation of a plane. Now we ask about the intersection of two planes, and almost none of them can do it. The scary part, is that very (very) few of the teachers of alg II and above that I have questioned about this could provide an answer either .
We begin by recalling that an equation in three variables, such as 2x+3y+z=6 can represent a plane in space. When students had three such equations that intersected in a unique point, they found the solution by one of several methods. Most students learn to solve such equations by the methods called elimination and substitution at the very least. Others may have also been introduced to Cramer’s rule for solving systems with determinants and perhaps two methods using matrices.

The most commonly taught matrix method is to write a matrix equation and then solve it using the inverse matrix method. A second, and as we will point out, more efficient and general method is the Gauss-Jordan reduced row-echelon form (RREF) of an augmented matrix. We give an example of both below to clarify the terminoligy.

We begin with three planes determined by the equations {x + y – 2z = 9; 2x – 3y + z = -2; and x + 3y + z = 2} This same system of equations can be expressed as the matrix equation.

Notice that the left matrix is made up of the coefficients of the three variable terms in each equation, and the right matrix contains the constant terms. We can find the intersection by taking the inverse of the left matrix and multiplying on the left of both sides of the equation. The simplified result gives

This seems to be the most commonly taught method, and the one that students and teachers seem to prefer, and yet it has two major disadvantages. The first disadvantage is that it tells you little or nothing about systems which have a solution, but not a single unique solution. In fact it most students (or teachers?) can not distinguish between the cases (and there are several different ones) with no solutions frorm the ones with an infinite number of solutons. This same defect applies to attempts to use Cramers Rule. The second problem is that the inverse method is more computationally complex, that is, it takes more operations for the solution than the alternative RREF method, and the difference grows as problems reach higher orders of magnitude. For the problems that are generally assigned at the high school level, the difference in computability presents no real problem, but the difference in the range of applicable questions can be very significant in a students understanding of general systems of three equations.

In contrast with the Inverse method that will only work if the three planes intersect in a single point, the RREF form will allow us to work with systems which do not even have the same number of equations as unknowns. This is the type of situation created when we try to find the line of intersection of two planes.

RREF for two planes

We will use the equations 2x + 3y – 3z = 14 and –3x + y + 10z = -32. When we write an augmented matrix for the system of only two equations we get a 2x4 matrix, shown here:

When we reduce this system, by matrices or otherwise, we get

which is a matrix expression of x - 3z = 10 and y + z = -2.

We notice that both equations contain a z variable, it might occur to us to ask, “What happens if we substitute different values in for z?”. For example, if we try z=0 we note that from the first equation we get x=10 and from the second we get y=-2. What does this tell us about the point (10, -2, 0). If we check it against the two original equations we notice that the point makes both equations true, 2(10) + 3(-2) – 3(0) = 14 and –3(10) + (-2) + 10(0) = -32. So the point (10, -2, 0) is on both planes and therefore must lie on the line that is their intersection.

Can we find more points? What happens if we try z=1 or z=2 or other values. Using z=1 we get x – 3(1) = 10 which simplifies to x=13 ; and y +(1) = -2 which simplifies to y=-3. Checking the point (13, -3, 1) we see that it also makes both equations true, and so it must also be on the line of intersection.

Writing the parametric equation of a line in three space

So now we have two points on the line of intersection; (10,-2, 0) and (13, -3, 1). How can we describe the line? One way is to write a formula for all the points so that someone could find as many values of (x,y,z) as they wish. We do this with parametric equations. A parametric equation is an equation that explains the values of one set of variables (in this case x, y, and z) in terms of another “parameter” which we will call t. We will define the line in terms of one point, and instructions to get from one point to another; sort of a three-space equivalent of slope.

When we worked with slope in a plane we had to find the change in x and the change in y, but now in three space we need a change in z also. If we look at our two known points, we can see that from the first we found, (10,-2, 0), to the second, (13, -3, 1), the x-value increases 3, the y-value decreases 1, and the z value increased 1. Let’s record those as a ordered triple, but to keep it separate from our points we will use brackets to enclose the changes; like this [3, -1, 1] .

We can even use this to find more points. If we add [3, -1, 1] to the last point we found (13, -3, 1) we get (16, -4, 2); and we know it is on both planes because it makes both the original equations true (Remember? They were 2x + 3y – 3z = 14 and –3x + y + 10z = -32) . Of course we do not have to add integer multiples of t, and if we allow t to be ANY real value, then we can write a general expression for all the points on the line of intersection in the form (x, y, z) = (10, -2, 0) + t [3, -1, 1] . [some texts will write this as a vector form (x, y, z) = (10+3t, -2 – t, t ) ].
And we have written the equation of the line of intersection. One of the nice things about the matrix approach is that it allows us to focus on interpreting the results... do three planes intersect in a line, in two parallel lines, in three lines all heading in the same direction? I will try to talk about these issues next, and then a little more about the equations of planes in space.
Post a Comment