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- 1. TARUNGEHLOT Systems of Equations The Geometry of three planes in spaceSome background Early in Geometry students learn that when two planes intersect, they intersect in asingle straight line. Few are taught the tools to determine the line of intersection, andexpress it with an equation. This paper attempts to provide an approach to developingthe tools to accomplish that task and expand their understanding of the informationavailable in the use of the reduced row-echelon form of the augmented matrix of a systemof three equations in three unknowns. We begin by recalling that an equation in three variables, such as 2x+3y+z=6 canrepresent a plane in space. When students had three such equations that intersected in aunique point, they found the solution by one of several methods. Most students learn tosolve 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 withdeterminants and perhaps two methods using matrices. The most commonly taught matrix method is to write a matrix equation and thensolve it using the inverse matrix method. A second, and as we will point out, moreefficient and general method is the Gauss-Jordan reduced row-echelon form (RREF) of anaugmented 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 thematrix equation 1 1 2 x 9 2 3 1 y 2 1 3 1 z 2 . Notice that the left matrix is made up of the coefficients of the three variable termsin each equation, and the right matrix contains the constant terms. We can find theintersection by taking the inverse of the left matrix and multiplying on the left of bothsides of the equation. The simplified result gives
- 2. 6 7 1 25 25 5 9 2 1 3 1 2 1 25 25 5 2 3 9 2 1 25 25 5This seems to be the most commonly taught method, and the one that students andteachers seem to prefer, and yet it has two major disadvantages. The first disadvantageis that it tells you little or nothing about systems which have a solution, but not a singleunique solution. The second is that the inverse method is more computationally complex,that is, it takes more operations for the solution than the alternative RREF method, andthe difference grows as problems reach higher orders of magnitude. For the problemsthat are generally assigned at the high school level, the difference in computabilitypresents no real problem, but the difference in the range of applicable questions can bevery significant in a students understanding of general systems of three equations. The first obvious case is three planes intersecting in a single point that is the mostfamiliar to students. In addition to the Inverse method shown above, a second matrixmethod using the augmented matrix to represent the system will also solve this system.This method takes a matrix that combines the coefficient matrix with the constant matrixand then uses the same row operations that students learned when they solved systemsby elimination to produce an augmented matrix with the solution easily apparent. Here is 1 1 2 9 2 3 1 2the augmented matrix for the system, 1 3 1 2 , and the RREF matrix of the same 1 0 0 2 0 1 0 1system 0 0 1 3 . In contrast with the Inverse method that will only work if the three planes intersectin a single point, the RREF form will allow us to work with systems which do not evenhave the same number of equations as unknowns. This is the type of situation createdwhen 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 wewrite an augmented matrix for the system of only two equations we get a 2x4 matrix, 2 3 3 14 3 1 10 32 .
- 3. 1 0 3 10 When we reduce this to the RREF we get 0 1 1 2 . Now how do we interpretthis result? If we convert the RREF of the augmented matrix back to a system ofequations we get 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 wenote that from the first equation we get x=10 and from the second we get y=-2. Whatdoes this tell us about the point (10, -2, 0). If we check it against the two originalequations we notice that the point makes both equations true, 2(10) + 3(-2) – 3(0) = 14and –3(10) + (-2) + 10(0) = -32. So the point (10, -2, 0) is on both planes and thereforemust 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 whichsimplifies to y=-3. Checking the point (13, -3, 1) we see that it also makes both equationstrue, 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 thatsomeone could find as many values of (x,y,z) as they wish. We do this with parametricequations. A parametric equation is an equation that explains the values of one set ofvariables (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 toanother; 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 thechange in y, but now in three space we need a change in z also. If we look at our twoknown 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’srecord those as a ordered triple, but to keep it separate from our points we will usebrackets 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 wefound (13, -3, 1) we get (16, -4, 2); and we know it is on both planes because it makesboth 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 beANY real value, then we can write a general expression for all the points on the line ofintersection in the form (x, y, z) = (10, -2, 0) + t [3, -1, 1] . [some texts will write this asa vector form (x, y, z) = (10+3t, -2 – t, t ) ]. If we return to examine the RREF form of the augmented matrix we wonder if the 1 0 3 10direction vector [3, -1, 1] is in anyway obvious from the reduced matrix 0 1 1 2 . Weare tantalized by the appearance of –3 and 1 in the Z column, and the 10, -2 in the
- 4. constants column. Could we have looked at this reduced form and gone directly to theequation of the line? Perhaps a second example will help. For the second two planes we pick { x+y + 5z = 8 and 2x + y + 8z = 13 }. Our 1 0 3 5reduced row echelon form looks like 0 1 2 3 . Is it possible that our line of intersection isas simple as (x, y, z) = (5,3,0) +t [-3, -2, 1] (clever folks are already thinking we couldwrite this last as [3, 2, -1] since t goes through all negative and positive values). Will thepoints generated lie on both planes. (5, 3, 0) is almost a trivial to check, and if we check(2, 1, 1) we see that it also is on both planes. We are now more assured that it is at leastplausible that the RREF matrix makes the intersection line simple to obtain.A second consistent solution to three planes Now let’s add to our system and make a system of three equations; {2x + 3y – 3z= 14 , –3x + y + 10z = -32, x + 7y + 4z = -4}. If we try to solve this system by theinverse method we get the dreaded “Singular Mat” error on the calculator and all we candetermine is that there is NOT a single unique solution. If we use the RREF we getsomething that should remind us of our previous example, and for good reason. BecauseI chose the third line to be a linear combination of the first two, the three planes all sharea common line of intersection, the line (x, y, z) = (10, -2, 0) + t [3, -1, 1] as shown by thereduced form: 1 0 3 10 0 1 1 2 0 0 0 0 The third row gives us an obviously true statement that reflects that the third equation,in essence, added no new information to the system. This “row of zeros” at the bottom isthe key to the student that the system had three equations but the intersection would becompletely determined by any two of them. When three planes intersect in space, there are five basic configurations that cantake place. I ask students to visualize and describe the five situations that can exist priorto discussion of the method of finding the solutions. In most years they come up with allfive configurations and names for them. Here is a list of the five as my students tend todescribe them. The two consistent systems are three planes intersecting in a single point, andthree planes intersecting in a single line (the book binding model – the planes are pages ina book and the common line is the binding). The three configurations with no common points of intersection are three parallelplanes, two parallel planes cut by a transverse plane (the H configuration) forming twoparallel lines, and three planes intersecting in mutual pairs to form three parallel lines (thedelta configuration). The first and the second are often easy to recognize from the
- 5. equations since the parallels are usually recognizable, but the third is difficult to identifywithout reducing.Three configurations w/o a common solution If we look at the three inconsistent systems possible with three planes, the moreobvious cases may lead us to more understanding of the less obvious cases. First I wantto look at three parallel planes, x + 2y + 3z = 1, x+ 2y + 3z = 5, and x + 2y + 3z = 10.The reduced RREF augmented matrix looks like this: 1 2 3 0 0 0 0 1 0 0 0 0 The first row looks like it is directly related to the coefficients of x, y, and z in eachequation, but the 0 in the constant term may seem perplexing. I think of this informationas a direction vector [1,2,3] that is perpendicular to the three parallel planes. The bottomline of zeros reminds us that the last equation added no new information about thesolution set. But what are we to make of the equation for the second row, 0x + 0y + 0z= 1. I tell students that this is the signal that there is NO common point of intersection.In essence, we have an intersection when 0=1, and that is never true. Now we examine a system with two of the same parallels and a third plane notparallel to them, the H configuration, { x + 2y + 3z = 1, x+ 2y + 3z = 5, 2x – y + z = 1}. 1 0 1 0 0 1 1 0The reduced matrix looks like 0 0 0 1 . How can we interpret this. The bottom row isour old insolvable friend 0x +0y + 0z = 1, reminding us that there can be no commonpoint on the three planes (since two of them are parallel). The upper two rows look likethe rows when we had a line of intersection except that both constant terms equal zero.Does this tell us anything about the two lines of intersection formed by the plane cuttingthrough the two parallel planes. Perhaps we can find out more if we find the twointersections one at a time. If we reduce the two equations x + 2y + 3z = 1, 2x – y + z = 1 0 1 .61, our reduced matrix looks like this, 0 1 1 .2 . This gives a line (x, y, z) = (.6, .2, 0) + t[-1, -1, 1] . If we do the intersection with the second parallel plane we get (x, y, z) =(1.4, 1.8, 0) + t [-1, -1, 1]. It seems that our original reduced system was flashing thedirection vector part of the two parallel lines created by the three planes in the thirdcolumn. For our final case, we pick the inconsistent situation which is most difficult todistinguish from the consistent cases, the delta configuration in which three planesintersect in mutual pairs to form three parallel lines. An example is the three planesdefined by { x + y + 7z = 5, x – y – z = 3, and 2x – y + 2z = 4} .
- 6. 1 0 3 0 0 1 4 0 The reduced form looks like this, 0 0 0 1 and we are now becoming accustom tothe 0001 on the bottom reminding us there is no common point for the three planes. Anda quick inspection of the three planes assures us that no two of them are parallel, so theonly possible case is the “delta formation” and three parallel lines. What are we to makeof the two upper rows. Our experience now leads us to think that [3,4,-1] will be thedirection vector for all three lines. We can verify our supposition by taking the vectors inpairs and reducing the two-equation systems to find the actual line of intersection, but asis the fashion in mathematical writing, I leave that final task as an exercise for the reader.

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