Intersection of planes

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2232221

1131211

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The solution will be one of three cases:

1. Exactly one solution, an ordered triple (x, y, z) 2. A dependent system with infinitely many solutions3. No solution

Three Equations Containing Three Variables

The first two cases are called consistent since there are solutions. The last case is called inconsistent.

Intersecting Planes

Any two planes that are not parallel or identical will intersect in a line and to find the line, solve the equations simultaneously.

For example in the figure above, the white plane and the yellow plane intersect along the blue line.

Example : Find the line of intersection for the planes x + 3y + 4z = 0 and x – 3y +2z = 0.

zyzy

zyxzyx

zyxzyx

31

or026

023023

0430431

Back substitute y into one of the first equations and solve for x.

zx

zzx

zzx

3

04

0431

3

Finally if you let z = t, the parametric equations for the line are tztytx

and

31

,3

Solution: To find the common intersection, solve the equations simultaneously. Multiply the first equation by –1 and add the two to eliminate x.

Planes intersect at a point: consistent with one solution

With two equations and two variables, the graphs were lines and the solution (if there was one) was where the lines intersected. Graphs of three variable equations are planes. Let’s look at different possibilities. Remember the solution would be where all three planes all intersect.

Planes intersect in a line: consistent system called dependent with an infinite number of solutions

Three parallel planes: no intersection so system called inconsistent with no solution

No common intersection of all three planes: inconsistent with no solution

We will be doing a “monster” elimination. (This just means it’s like elimination that you learned with two equations and variables but it’s now “monster” because the problems are bigger and meaner and uglier).

732

10223

42

zyx

zyx

zyx

Your first strategy would be to choose one equation to keep that has all 3 variables, but then use that equation to “eliminate” a variable out of the other two. I’m going to choose the last equation to “keep” because it has just x.

coefficient is a 1 here so will easy to work with

27)32(2 zyx

732

10223

42

zyx

zyx

zyx 732 zyxkeep over here for later use

Now use the third equation multiplied through by whatever it takes to eliminate the x term from the first equation and add these two equations together. In this case when added to eliminate the x’s you’d need a −2.

1055 zy

1055 zy

put this equation up with the other one we kept

14642 zyx42 zyx

37323 zyx

732

10223

42

zyx

zyx

zyx 732 zyx

21963 zyx10223 zyx

1174 zy

1055 zy

we won’t “keep” this equation, but we’ll use it together with the one we “kept” with y and z in it to eliminate the y’s.

Now use the third equation multiplied through by whatever it takes to eliminate the x term from the second equation and add these two equations together. In this case when added to eliminate the x’s you’d need a 3.

So we’ll now eliminate y’s from the 2 equations in y and z that we’ve obtained by multiplying the first by 4 and the second by 5

732

10223

42

zyx

zyx

zyx 732 zyxkeep over here for later use

1174 zy

1055 zy

we can add this to the one’s we’ve kept up in the corner

1055 zy 402020 zy553520 zy

1515 z1z

1z

Now we are ready to take the equations in the corner and “back substitute” using the equation at the bottom and substituting it into the equation above to find y.

732

10223

42

zyx

zyx

zyx 732 zyxkeep over here for later use1055 zy

Now we know both y and z we can sub them in the first equation and find x

10155 y

1z

55 y 1y

71312 x

75 x 2x

These planes intersect at a point, namely the point (2 , −1 , 1).The equations then have this unique solution. This is the ONLY x, y and z that make all 3 equations true.

I’m going to “keep” this one since it will be easy to use to eliminate x’s from others.

Let’s do another one:

143

52

032

zyx

zyx

zyx52 zyx

032

10242

zyx

zyx

10 zy

10 zy

we’ll keep this one

143

15363

zyx

zyx

1622 zy

Now we’ll use the 2 equations we have with y and z to eliminate the y’s.

If we multiply the equation we kept by 2 and add it to the first equation we can eliminate x’s.

If we multiply the equation we kept by 3 and add it to the last equation we can eliminate x’s.

143

52

032

zyx

zyx

zyx 52 zyx

10 zy

10 zy

we’ll multiply the first equation by -2 and add these together

1622 zy

Oops---we eliminated the y’s alright but the z’s ended up being eliminated too and we got a false equation.

2022 zy1622 zy

40

This means the equations are inconsistent and have no solution. The planes don’t have a common intersection and there is not any (x, y, z) that make all 3 equations true.

Let’s do another one:

454

22

12

zyx

zyx

zyx I’m going to “keep” this one since it will be easy to use to eliminate x’s from others.

12 zyx

22

2422

zyx

zyx

033 zywe’ll keep this one

454

4844

zyx

zyx

If we multiply the equation we kept by −2 and add it to the second equation we can eliminate x’s.

033 zy

Now we’ll use the 2 equations we have with y and z to eliminate the y’s.

033 zyIf we multiply the equation we kept by −4 and add it to the last equation we can eliminate x’s.

454

22

12

zyx

zyx

zyx 12 zyx

033 zy

multiply the first equation by −1 and add the equations to eliminate y.

033 zy

033 zy

033 zy033 zy

00

Oops---we eliminated the y’s alright but the z’s ended up being eliminated too but this time we got a true equation.

This means the equations are consistent and have infinitely many solutions. The planes intersect in a line. To find points on the line we can solve the 2 equations we saved for x and y in terms of z.

454

22

12

zyx

zyx

zyx 12 zyx033 zyzz

First we just put z = z since it can be any real number. Now solve for y in terms of z.

033 zy

zy

Now sub it −z for y in first equation and solve for x in terms of z.

12 zzxzx 1

The solution is (1 − z , −z , z) where z is any real number.

For example: Let z be 1. Then (0 , −1 , 1) would be a solution.Notice is works in all 3 equations. But so would the point

you get when z = 2 or 3 or any other real number so there are infinitely many solutions.

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au