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This Lesson (LINEAR EQUATIONS IN THREE DIMENSIONS) was created by by Theo(13342)
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This lesson covers an overview of LINEAR EQUATIONS IN THREE DIMENSIONS.
REFERENCES There are multiple references and there is considerable overlap between references. Select the references you find easier to understand first. Once you have developed some expertise, you can then go back to the more difficult references if you find the need to do so. This lesson is an overview of at least some of the material in the references, though not necessarily all. Linear Equations in Three Variables (wtamu) Linear Systems in Three Variables (lamar) Systems of Equations in Three Variables (sosmath) How to solve Systems of Linear Equations in Three Variables (ehow) Description of Linear Equations with [One,Two,Three] Dimensions Solving Linear Equations Basics Review (purplemath) Systems of Linear Equations in three variables (cliffsnotes) Systems of Equations in Three Variables (sosmath) GRAPH OF THREE DIMENSIONS A two dimensional graph has height and width, but no depth. The height is usually represented by the y-axis and the width is usually represented by the x-axis. A three dimensional graph has height and width and depth. The height is usually represented by the z-axis, the depth is usually represented by the x-axis, and the width is usually represented by the y-axis. A picture of a two dimensional graph grid next to a three dimension graph grid can be seen by clicking on the following hyperlink. Two and Three Dimensional Graph Grids Most three dimensional graphing programs have a rotating feature to allow you to see the graph at various angles. This is essential as three dimensional objects can be difficult to visualize completely without this feature. The only thing you need to watch for is to keep track of where the x-axis and the y-axis and the z-axis are in relation to each other. Click on the following hyperlink to see a picture of the true zero point of the three dimensional graph. Zero Points of Three Dimensional Graph SOLVING SYSTEMS OF LINEAR EQUATIONS IN THREE DIMENSIONS Linear Equations in three dimensions are equations of planes. This is different from Linear Equations in two dimensions. Linear equations in two dimensions are straight lines. The standard form of linear equations in three dimensions is: ax + by + cz = d The standard form of linear equations in two dimensions is: ax + by = c There is the addition of the third variable z that creates the third dimension. This third variable is what causes the equation to form a plane, rather than a line. Click on the following hyperlink to see a graphical presentation of the two dimensional equation of y = x next to the three dimensional equation of z = x + y. Two and Three Dimensional Graphs Planes can intersect in either a line or a point. If they do not intersect at all, then they are parallel to each other. If they intersect everywhere, then they are identical to each other which means they are the same plane. In two dimensional space, lines intersect in a point. If they do not intersect at all, then they are parallel to each other. If they intersect everywhere, then they are identical to each other which means they are the same line. Three dimensional space is just an extension of two dimensional space in this regard. There is the addition of a third variable which is what adds the third dimension to the equations. Click on the following hyperlink to see examples of planes that are parallel to each other and planes that intersect in a point or a line. Intersecting and Parallel Planes If you have three equations of planes and the planes intersect in a point, then you will have a solution for your system of equations that will provide a unique value for each of the three unknowns. We will be working with x,y,z variable names even though other variable names could be used. For graphing purposes, x,y,z variable names are necessary in order to satisfy the requirements of most graphing software. SOLVING SYSTEMS OF EQUATIONS IN THREE DIMENSIONS BY THE ELIMINATION METHOD The basic method is as follows: Note that numbering equations is for referencing purposes. I do this to make it easy to tell you what equation to work on. You can do it for yourself as well, but it is not a requirement. Number your equations 1, 2, and 3. Take equations 1 and 2 and eliminate the z variable. Name the resulting equation from this as equation 4. Take equations 2 and 3 and eliminate the z variable. Name the resulting equation from this as equation 5. Take equations 4 and 5 and eliminate the y variable. Solve for x in the remaining equation. You could number this equation 6 but it is not necessary. Replace x with the value of x and solve for y in equation 4. Replace x and y with the value of x and y in equation 1 and solve for z. Replace x and y and z in equations 1, 2, and 3 in order to confirm that your results are valid for all three equations. They must be valid for all three equations or the values have not solved all three equations simultaneously. Note that you could have solved for y in equation 4 or 5. I specified equation 4 to keep to the systematic approach. Note also that you could have solved for z in equation 1, 2, or 3. I specified equation 1 to keep to the systematic approach. If there is a unique value for x and y and z, then you will find it this way. SOLVING SYSTEMS OF EQUATIONS IN THREE DIMENSIONS BY THE SUBSTITUTION METHOD The basic method is as follows: Number your equations 1, 2, and 3. Take equation 1 and solve for z in terms of x and y. Substitute for z in equation 2 and solve for y in terms of x. Substitute for z in equation 3 and then substitute for y in equation 3 and then solve for x. Substitute the value of x from equation 3 and solve for y in equation 2. Substitute the value of x from equation 3 and the value of y from equation 2 and solve for z in equation 1. Replace x and y and z in equations 1, 2, and 3 in order to confirm that your results are valid for all three equations. Note that this is the systematic approach again. You could mix up the order of the equations you used to solve for each variable. The main idea is that you use a different equation each time to avoid finding a solution for one of the equations that would not be applicable to the others. The systematic approach avoids confusion as to which equations you used and which you have left to use. SOLVING SYSTEMS OF EQUATIONS IN THREE DIMENSIONS BY USING BOTH METHODS WHERE APPROPRIATE. With this combined method, you start off with the elimination method and substitute where appropriate if it makes subsequent calculations easier. I will use this method. Elimination will be the main method, but substitution will be used where it makes sense to do so. There are other more sophisticated methods that will be covered in separate lessons. These are generally based on the elimination method being shown here. ORDER OF COMPUTATIONS I have found that going down the line systematically is better in the long run than jumping around to find which equation to calculate first, then which variable to eliminate first, etc. What often happens is that when you grab the equations that look simpler to calculate, you wind up with equations that have complications down the road. You also lose track of which equation you did and which you didn't. The systematic approach makes the process of getting the answer simpler even though it might not seem so at first. That is why I compared equations 1 and 2 first, then 2 and 3, in that order. It is also why I eliminated z first, then y, then solved for x, then solved for y, etc. You are free to use whatever method works best for you. Don't feel constrained just because I did it this way. I prefer to do the problems this way. There is the additional benefit of making it easier to present them to you this way as well. If I don't have to present the problems, then I might ake some shortcuts once I get enough experience in solving the problems and learn what those shortcuts are. The same will happen for you as you gain more experience in solving these problems. MISCELLANEOUS INFORMATION REGARDING LINEAR EQUATIONS IN THREE DIMENSIONS If all three variables disappear from the resulting equation when you compare two of the equations with each other, and the equation is true, then this means that the planes represented by these equations are identical which means they are the same plane. If all three variables disappear from the resulting equation when you compare two of the equations with each other, and the equation is false, then this means that the planes represented by these equations are parallel to each other. If you eliminate one of the variables and are working on eliminating the other two, and all two variables disappear from the resulting equation, and the equation is true, this means that the lines are identical. If they are identical, this means they are the same line which means that the planes you created these equations from are all intersecting in a line and not a point. If you eliminate one of the variables and are working on eliminating the other two, and all two variables disappear from the resulting equation, and the equation is false, this means that the lines are parallel. If the lines are parallel, this means that two of the planes you created these equations from are parallel to each other and the third plane intersects with each of them in a line. SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 1 This system of equations will yield a unique solution for each variable because all three planes intersect in a point. The equations are: x + y - z = -2 (equation 1) x - y - z = 0 (equation 2) x - y + z = 0 (equation 3) First we take equation 1 and equation 2 and eliminate the z variable. x + y - z = -2 (equation 1) x - y - z = 0 (equation 2) Since the z variables have a common coefficient, we don't have to do any multiplying or dividing to get one. Since the signs of the z variables are the same, we will subtract equation 2 from equation 1 in order to eliminate them. x + y - z = -2 (equation 1) minus: x - y - z = 0 (equation 2) equals: 0 + 2y = -2 (equation 4) Two of the variables dropped out, rather than 1. We solve for y and we get: y = -1 (equation 4) This has been named equation 4 for future reference. Next we take equation 2 and 3 and eliminate the z variable. x - y - z = 0 (equation 2) x - y + z = 0 (equation 3) Since the z variables have a common coefficient, we don't have to multiply or divide either equation to get a common coefficient. Since the signs are opposite, we will add equation 3 to equation 2 in order to eliminate them. x - y - z = 0 (equation 2) plus: x - y + z = 0 (equation 3) equals: 2x - 2y = 0 (equation 5) This has been named equation 5 for future reference. Next we take equation 4 and 5 and eliminate the y variable. y = -1 (equation 4) 2x - 2y = 0 (equation 5) Since y has already been solved for in equation 4, we will skip eliminating y and use it to solve for x in equation 5. 2x - 2y = 0 (equation 5) replace y with -1 to get: 2x - 2*(-1) = 0 simplify to get: 2x + 2 = 0 subtract 2 from both sides of the equation to get: 2x = -2 divide both sides of the equation by 2 to get: x = -1 We have two of the variables we need. They are: x = -1 y = -1 We take equation 1 and substitute for x and y to solve for z. x + y - z = -2 (equation 1) replace x with -1 and replace y with -1 to get: -1 + (-1) - z = -2 simplify to get: -2 - z = -2 add 2 to both sides of the equation to get: -z = 0 multiply both sides of the equation by -1 to get: z = 0 We have all three variables. They are: x = -1 y = -1 z = 0 We take the three original equations and replace x and y and z with these solutions to confirm that they will solve all three equations simultaneously. The three original equations are: x + y - z = -2 (equation 1) x - y - z = 0 (equation 2) x - y + z = 0 (equation 3) replace x with -1 and y with -1 and z with 0 to get: (-1) + (-1) - 0 = -2 (equation 1) (-1) - (-1) - 0 = 0 (equation 2) (-1) - (-1) + 0 = 0 (equation 3) simplify and combine to get: -2 = -2 (equation 1) 0 = 0 (equation 2) 0 = 0 (equation 3) All three equations are true confirming that the solutions we calculated are good. Click on the following hyperlink to see a graph of the solution to this equation. The graph shows that the three planes intersect in a point at (x,y,z) = (-1,-1,0). Solving Linear Equations in 3 Dimensions Example 1 SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 2 This is the same as example 1. The difference here is that we will be solving the equations by substitution rather than elimination. The technique of substitution uses equation 1 to solve for z in terms of x and y. It then uses equation 2 to solve for y in terms of x. It then uses equation 3 to solve for x. It then goes back to equation 2 and solves for y. It then goes back to equation 1 and solves for z. It then replaces x, y, and z with the values for x, y, and z in all three equations to confirm that the answer applies to all of the equations simultaneously. The equations are: x + y - z = -2 (equation 1) x - y - z = 0 (equation 2) x - y + z = 0 (equation 3) First we take equation 1 and solve for z in terms of x and y. x + y - z = -2 (equation 1) add z to both sides of this equation and add 2 to both sides of this equation. z = x + y + 2 (equation 4) This has been named equation 4 for future reference. Next we take equation 2 and replace z with the value obtained in equation 4 and solve for y in terms of x. x - y - z = 0 (equation 2) replace z with (x+y+2) from equation 4 to get: x - y - (x + y + 2) = 0 simplify to get: x - y - x - y - 2 = 0 simplify and combine like terms to get: -2y - 2 = 0 add 2y to both sides of the equation to get: 2y = -2 divide both sides of the equation by 2 to get: y = -1 (equation 5) This has been named equation 5 for future reference. Next we take equation 3 replace z with the value obtained in equation 4 and replace y with the value obtained in equation 5 and solve for x. x - y + z = 0 (equation 3) replace z with (x+y+2) from equation 4 to get: x - y + (x+y+2) = 0 remove parentheses to get: x - y + x + y + 2 = 0 simplify and combine like terms to get: 2x + 2 = 0 subtract 2 from both sides of the equation to get: 2x = -2 divide both sides of the equation by 2 to get: x = -1 (equation 6) Note that we did not have to replace y with the value obtained from equation 5 because y canceled out of the equation before we needed to do so. This has been named equation 6 for future reference. Next we would go to equation 2 and solve for y, but since we already have a solution for y from equation 5, we don't really need to go back and solve for it again. We'll skip going back to equation 2 to solve for y and go directly to equation 1 to solve for z. x + y - z = -2 (equation 1) replace x with -1 from equation 6 and replace y with -1 from equation 5 to get: -1 + -1 - z = -2 simplify and combine like terms to get: -2 -z = -2 add 2 to both sides of this equation to get: -z = 0 multiply both sides of this equation by (-1) to get: z = 0 We have solutions for this system of equation as follows: x = -1 y = -1 z = 0 Since this it the same solution as we got through elimination in example 1, we do not need to go any further to confirm that the results are good. SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 3 This example will also result in a common solution for all three equations that will be the intersection of three planes in a point graphically. This example is a little more complex but the method is the same. We will use the method of elimination from now on, only using substitution where it makes sense to do so. Our system of equations for example 3 are: 5x + 3y - 2z = 0 (equation 1) x + 2y + 3z = 5 (equation 2) 2x - 2y + 4z = -5 (equation 3) We take equation 1 and 2 and solve to eliminate z. 5x + 3y - 2z = 0 (equation 1) x + 2y + 3z = 5 (equation 2) Multiply equation 1 by 3 and multiply equation 2 by 2 to get: 15x + 9y - 6z = 0 (equation 1 multiplied by 3) 2x + 4y + 6z = 10 (equation 2 multiplied by 2) We now have common coefficients for z. Since the signs are different, we will add equation 2 to equation 1 to get: 15x + 9y - 6z = 0 (equation 1) plus: 2x + 4y + 6z = 10 (equation 2) equals: 17x + 13y = 10 (equation 4) This has been named equation 4 for future reference. We then take equation 2 and 3 and solve to eliminate z. x + 2y + 3z = 5 (equation 2) 2x - 2y + 4z = -5 (equation 3) Multiply equation 2 by 4 and multiply equation 3 by 3 to get: 4x + 8y + 12z = 20 (equation 2 multiplied by 4) 6x - 6y + 12z = -15 (equation 3 multiplied by 3) We now have common coefficients for z with the same sign. Because the signs are the same, we will subtract equation 3 from equation 2. 4x + 8y + 12z = 20 (equation 2) minus: 6x - 6y + 12z = -15 (equation 3) equals: -2x + 14y = 35 (equation 5) This has been named equation 5 for future reference. We have eliminated z in equation 4 and equation 5. We now take equation 4 and equation 5 and solve to eliminate the y variable. 17x + 13y = 10 (equation 4) -2x + 14y = 35 (equation 5) We multiply equation 4 by 14 and we multiply equation 5 by 13 to get: 238x + 182y = 140 (equation 4 multiplied by 14) -26x + 182y = 455 (equation 5 multiplied by 13) We now have common coefficients for y with the same signs. Because the signs are the same, we will subtract equation 5 from equation 4. 238x + 182y = 140 (equation 4) minus: -26x + 182y = 455 (equation 5) equals: 264x = -315 divide both sides of this equation by 264 to get: x = -315/264 = -1.193181818 We go back to the original equation 4 and solve for y. 17x + 13y = 10 (equation 4) replace x with -1.193181818 to get: 17*(-1.193181818) + 13y = 10 simplify to get: -20.28409091 + 13y = 10 add 20.28409091 to both sides of this equation to get: 13y = 30.28409091 divide both sides of this equation by 13 to get: y = 2.329545455 We have so far: x = -1.193181818 y = 2.329545455 We now go back to equation 1 and solve for z. 5x + 3y - 2z = 0 (equation 1) replace x with -1.193181818 and replace y with 2.329545455 to get: 5*(-1.193181818) + 3*(2.329545455) - 2z = 0 simplify and combine like terms to get: 1.022727273 - 2z = 0 add 2z to both sides of this equation to get: 2z = 1.022727273 divide both sides of this equation by 2 to get: z = .511363636 We now have: x = -1.193181818 y = 2.329545455 z = .511363636 We now go back to the three original equations and replace x and y and z with our solutions to confirm that these solutions are valid for all three equations. 5x + 3y - 2z = 0 (equation 1) x + 2y + 3z = 5 (equation 2) 2x - 2y + 4z = -5 (equation 3) replace x with -1.193181818 and y with 2.329545455 and z with .511363636 to get: 5*(-1.193181818) + 3*2.329545455 - 2*.511363636 = 0 (equation 1) -1.193181818 + 2*2.329545455 + 3*.511363636 = 5 (equation 2) 2*(-1.193181818 - 2*2.329545455 + 4*.511363636 = -5 (equation 3) simplify and combine like terms to get: 0 = 0 (equation 1) 5 = 5 (equation 2) -5 = -5 (equation 3) The values calculated for (x,y,z) are good because they solve all three equations simultaneously. They are: x = -1.193181818. y = 2.329545455. z = .511363636. A graph of the planes created by this system of equations can be found by clicking on the following hyperlink. Solving Linear Equations in 3 Dimensions Example 3 IDENTICAL PLANES The standard form of linear equations in three dimensions is: ax + by + cz = d If all of the terms in the standard form of the equations of the planes are exact multiples of each other by the same factor, then the planes created by these equations are identical to each other. This means that they represent the same plane. Example: x + y + z = 3 2x + 2y + 2z = 6 3x + 3y + 3z = 9 These planes are identical to each other. All of the terms in the second equations are 2 times the corresponding terms in the first equation. All of the terms in the third equation are 3 times the corresponding terms in the first equation. To graph these planes, you have to solve for z. Once you do that, you will find that the equations are identical. First equation becomes: z = - x - y + 3 Second equation becomes: 2z = - 2x - 2y + 6 divide both sides of this equation by 2 to get: z = - x - y + 3 which is identical to the first equation. Third equation becomes: 3z = - 3x - 3y + 9 divide both sides of this equation by 3 to get: z = - x - y + 3 which is identical to the first and the second equation. PARALLEL PLANES The standard form of linear equations in three dimensions is: ax + by + cz = d If all of the terms in the standard form of the equations of the planes are exact multiples of each other by the same factor, except for the constant d term, then the planes are parallel to each other. This means that they will never intersect with each other. Example: x + y + z = 3 (equation 1) 2x + 2y + 2z = 12 (equation 2) 3x + 3y + 3z = 27 (equation 3) To graph these equations, solve for z. Equations become: z = - x - y + 3 (equation 1) z = - x - y + 6 (equation 2) z = - x - y + 9 (equation 3) A graph of these equations can be seen by clicking on the following hyperlink. Parallel Planes SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 4 This example will show what happens when all planes created by the system of equations are identical. Your equations are: x + y + z = 3 (equation 1) 2x + 2y + 2z = 6 (equation 2) 3x + 3y + 3z = 9 (equation 3) Take equation 1 and 2 and eliminate the z variable. x + y + z = 3 (equation 1) 2x + 2y + 2z = 6 (equation 2) Multiply equation 1 by 2 to get: 2x + 2y + 2z = 6 (equation 1 multiplied by 2) 2x + 2y + 2z = 6 (equation 2) Subtract equation 2 from equation 1 to get: 2x + 2y + 2z = 6 (equation 1) minus: 2x + 2y + 2z = 6 (equation 2) equals: 0 + 0 + 0 = 0 All three variables have dropped out and the equation is true because 0 equals 0. This means that equations 1 and 2 are identical which means they represent the same plane. Take equation 2 and 3 and eliminate the z variable. 2x + 2y + 2z = 6 (equation 2) 3x + 3y + 3z = 9 (equation 3) Multiply equation 2 by 3 and multiply equation 3 by 2 to get: 6x + 6y + 6z = 18 (equation 2 multiplied by 3) 6x + 6y + 6z = 18 (equation 3 multiplied by 2) Subtract equation 3 from equation 2 to get: 6x + 6y + 6z = 18 (equation 2) minus: 6x + 6y + 6z = 18 (equation 3) equals: 0 + 0 + 0 = 0 All three variables have dropped out and the equation is true because 0 equals 0. This means that equations 2 and 3 are identical which means they represent the same plane. Since equations 1 and 2 are identical, and equations 2 and 3 are identical, then all three equations are identical which means that they all represent the same plane. Your soluton is that the planes are all identical. SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 5 This example will show you what happens when all planes created by the system of equations are parallel. Your equations are: x + y + z = 2 (equation 1) 2x + 2y + 2z = 8 (equation 2) 3x + 3y + 3z = 18 (equation 3) Take equation 1 and 2 and solve to eliminate z. x + y + z = 2 (equation 1) 2x + 2y + 2z = 8 (equation 2) Multiply equation 1 by 2 to get: 2x + 2y + 2z = 4 (equation 1 multiplied by 2) 2x + 2y + 2z = 8 (equation 2) Subtract equation 2 from equation 1 to get: 0 + 0 + 0 = -4 All the variables dropped out and the equation is false. This occurs when the planes are parallel to each other. This means that the planes created by equations 1 and 2 are parallel to each other. Take equation 2 and 3 and solve to eliminate z. 2x + 2y + 2z = 8 (equation 2) 3x + 3y + 3z = 18 (equation 3) Multiply equation 2 by 3 and multiply equation 3 by 2 to get: 6x + 6y + 6z = 24 (equation 2 multiplied by 3) 6x + 6y + 6z = 36 (equation 3 multiplied by 2) Subtract equation 3 from equation 2 to get: 0 + 0 + 0 = -12 All the variables dropped out and the equation is false. This occurs when the planes are parallel. to each other. This means that the planes created by equations 2 and 3 are parallel to each other. Planes created by equations 1 and 2 are parallel, and planes created by equations 2 and 3 are parallel. This means that all the planes are parallel to each other. Your solution to this problem is that there is no solution. Click on the following hyperlink to see a graph of these equations. Solving Linear Equations in three dimensions Example 5 SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 6 Sometimes your planes will all intersect in a line. When this happens, you should be able to eliminate one of the variables successfully. When you try to eliminate the second variable, both remaining variables should drop out. You should be left with equations of lines that are identical. Your three equations are: x + y - z = 0 (equation 1) x - y - z = 0 (equation 2) y = 0 (equation 3) The first thing you notice is that equation 3 has already solved for y. this allows you to automatically reduce your system of equations from three equations in three unknowns to two equations in two unknowns. You take equation 1 and 2 and you replace y with 0. x + y - z = 0 (equation 1) x - y - z = 0 (equation 2) Replace y with 0 to get: x + 0 - z = 0 (equation 1) x - 0 - z = 0 (equation 2) Simplify to get: x - z = 0 (equation 4) x - z = 0 (equation 5) I will assign equation 4 to the modified equation 1 to distinguish it from equation 1. I will assign equation 5 to the modified equation 2 to distinguish it from equation 2. Subtract equation 5 from equation 4 to get: 0 = 0 Both variables drop out and the equation is true. This means that the equations are identical. This means that the lines created by these equations are the same line. This means that the planes that created these equations intersect in the same line. Since equations 4 and 5 are identical, then solving one of the equations automatically solves the other. Solve for x in equation 4. x - z = 0 (equation 4) add z to both side of this equation. x = z the equation of the line of intersection for these planes is x = z. The value of y will be 0 from the original equation 3. x and z can be any value as long as y = 0 and x = z. A graph of these equations can be seen by clicking on the following hyperlink. Solving Linear Equations in three dimensions Example 6 SOLVING LINEAR EQUATIONS IN THREE DIMENSIONS EXAMPLE 7 Sometimes two planes will be parallel to each other and intersect with a third plane in a line. The following equation shows such a situation. 2x + 3y - z = 0 (equation 1) 4x + 6y - 2z = 20 (equation 2) .5x - .5y - z = 2 (equation 3) We take equation 1 and 2 and eliminate the z variable. 2x + 3y - z = 0 (equation 1) 4x + 6y - 2z = 20 (equation 2) Multiply equation 1 by two to get: 4x + 6y - 2z = 0 (equation 1 multiplied by 2) 4x + 6y - 2z = 20 (equation 2) Since the coefficients of z are now the same terms and their signs are the same, we will subtract equation 2 from equation 1. 4x + 6y - 2z = 0 (equation 1) minus: 4x + 6y - 2z = 20 (equation 2) equals: 0 + 0 + 0 = -20 (equation 4) All the variables dropped out and the equation is false indicating there is no solutions between these two equations indicating that planes created by these equations are parallel to each other. At this point you can assume that there will not be a point solution or a line solution to all three equations at the same time since two of the planes are parallel to each other. The best that could happen is the third plane intersects with the two parallel planes in a line. You can stop here and just say there is no solution because two of the planes are parallel to each other. We will , however, continue further just to show you how this ends up. Next we continue in our systematic approach to compare equation 2 to equation 3 to eliminate z. 4x + 6y - 2z = 20 (equation 2) .5x - .5y - z = 2 (equation 3) Multiply equation 3 by 2 4x + 6y - 2z = 20 (equation 2) x - y - 2z = 4 (equation 3) Since the z term coefficients are the same and their signs are also the same, subtract equation 3 from equation 2 to eliminate the z. 4x + 6y - 2z = 20 (equation 2) minus: x - y - 2z = 4 (equation 3) equals: 3x + 7y = 16 (equation 4) This has been named equation 4 for future reference. Next compare equation 1 to equation 3 to eliminate z. 2x + 3y - z = 0 (equation 1) .5x - .5y - z = 2 (equation 3) Since the coefficients of the z term are the same and their signs are the same, subtract equation 3 from equation 1 to eliminate z. 2x + 3y - z = 0 (equation 1) minus: .5x - .5y - z = 2 (equation 3) equals: 1.5x + 3.5y = -2 (equation 5) This has been named equation 5 for future reference Compare equation 4 with equation 5 to eliminate y. 3x + 7y = 16 (equation 4) 1.5x + 3.5y = -2 (equation 5) Multiply equation 5 by 2 to get: 3x + 7y = 16 (equation 4) 3x + 7y = -4 (equation 5 multiplied by 2) Subtract equation 5 from equation 4 to eliminate z. 3x + 7y = 16 (equation 4) minus: 3x + 7y = -4 (equation 5) equals: 0 + 0 = 20 Both variables drop out and the equation is false indicating that the two lines created by these equations are parallel. You have two parallel planes with a third plane intersecting with each of them in a straight line. The two straight lines are parallel to each other. Click on the following hyperlink to see a graph of these two parallel planes intersecting with the third plane. Solving Linear Equations in three dimensions Example 7 This concludes the lesson on Linear Equations in Three Dimensions. This lesson has been accessed 20762 times. |
