Grade by Grade
C. Geometry and Algebra of Vectors · Lines and Planes

Solving Systems of Three Planes Algebraically

C4.4 C3.3About 75 minutes
Learning goals
  • I can solve a system of three linear equations in three unknowns by elimination and back-substitution.
  • I can recognize from the algebra when a system has one solution, infinitely many solutions, or none.
  • I can write infinitely many solutions using one or two parameters.
  • I can connect the algebraic result to the geometric configuration of the three planes.

Words to know

elimination
Adding or subtracting multiples of equations to remove a variable.
back-substitution
Putting known values into earlier equations to find the remaining variables.
inconsistent
Having no solution; elimination produces a false statement such as 0 = 3.
dependent (redundant) equation
An equation that is a combination of the others; elimination reduces it to 0 = 0.
Gaussian elimination
A systematic, row-by-row form of elimination used in linear algebra.

From pictures to answers

In the last lesson you sorted three planes into eight pictures. Now you will compute the intersection. The tool is elimination, the same method you used for two equations in Grade 10, extended to three unknowns. It always works, and the way it ends tells you which picture you are looking at.

This is exactly the method (called Gaussian elimination when it is organized in rows) that first-year university linear algebra uses for systems with many more unknowns. Engineers, economists and computer graphics programmers solve such systems constantly.

The method

  1. Label the equations (1), (2), (3).
  2. Choose a variable to eliminate, often one with coefficient 1 or −1.
  3. Combine (1) with (2), and (1) or (2) with (3), to remove that variable. You now have two equations in two unknowns: (4) and (5).
  4. Combine (4) and (5) to eliminate a second variable. Solve for the remaining one.
  5. Back-substitute to find the other two variables.
  6. Check the point in all three original equations.

Reading the result

  • Every variable has one value: the planes meet in one point.
  • A true statement 0 = 0 appears once: one equation was redundant. Let one variable be a parameter t; the planes share a line.
  • 0 = 0 twice (all three equations are multiples of one another): the planes coincide in a whole plane; use two parameters.
  • A false statement 0 = k (k ≠ 0): the system is inconsistent, with no solution. Check the normals: if two are parallel, you have parallel planes (with or without a third crossing them); if none are parallel, you have a triangular prism.
Start: three planesx + y + z = 62x − y + z = 3x + 2y − z = 2Eliminate z2x + 3y = 83x + y = 5Solve and back-substitutex = 1, y = 2z = 6 − 1 − 2 = 3(1) + (3) and (2) + (3)remove z from two equationsone point: (1, 2, 3)
Elimination turns three equations in three unknowns into two equations in two unknowns, then one in one. Back-substitution finishes the job. Here the planes meet at the single point (1, 2, 3).
After elimination you get…x, y, z all foundone point0 = 0 oncea line (1 parameter)0 = 0 twicea plane (2 parameters)0 = k, k ≠ 0no solutiontwo normals parallel?yes: parallel planesno: triangular prism
How to read the end of an elimination. A true statement 0 = 0 means an equation was redundant. A false statement 0 = k means the system is inconsistent; then the normals tell you which "no solution" picture you have.
Worked example

Example 1: One point

Solve x + y + z = 6 (1), 2x − y + z = 3 (2), x + 2y − z = 2 (3).

  1. Eliminate z. (1) + (3): 2x + 3y = 8 (4). (2) + (3): 3x + y = 5 (5).
  2. From (5), y = 5 − 3x. Substitute into (4): 2x + 15 − 9x = 8, so −7x = −7 and x = 1.
  3. Then y = 5 − 3 = 2, and from (1), z = 6 − 1 − 2 = 3.
  4. Check: (2) 2 − 2 + 3 = 3 ✓; (3) 1 + 4 − 3 = 2 ✓.

The three planes meet at the single point (1, 2, 3).

Worked example

Example 2: A common line

Solve x + y + z = 2 (1), x − y + 2z = 1 (2), 3x + y + 4z = 5 (3).

  1. Eliminate y. (1) + (2): 2x + 3z = 3 (4). (3) − (1): 2x + 3z = 3 (5).
  2. (5) − (4): 0 = 0. True, so one equation is redundant (indeed (3) = 2 × (1) + (2)).
  3. Let z = t. From (4), x = (3 − 3t)/2. From (1), y = 2 − x − z = (1 + t)/2.
  4. Choosing t = 1 gives the point (0, 1, 1). Check: 0 + 1 + 1 = 2 ✓, 0 − 1 + 2 = 1 ✓, 0 + 1 + 4 = 5 ✓.
  5. The direction (multiply the t-coefficients −3/2, 1/2, 1 by 2) is [−3, 1, 2].

The planes share the line r = [0, 1, 1] + s[−3, 1, 2]. No two normals are parallel, so this is three planes through one line, like book pages.

Worked example

Example 3: No solution

Change the last equation of Example 2 to 3x + y + 4z = 7 and solve.

  1. (1) + (2) still gives 2x + 3z = 3. Now (3) − (1) gives 2x + 3z = 5.
  2. Subtract: 0 = 2. False.
  3. The normals [1, 1, 1], [1, −1, 2], [3, 1, 4] are not parallel in pairs.

No solution. The planes form a triangular prism: each pair meets in a line, but no point is on all three.

Worked example

Example 4: A whole plane

Solve x − 2y + z = 3, 2x − 4y + 2z = 6, −x + 2y − z = −3.

  1. The second equation is 2 times the first, and the third is −1 times the first. Elimination gives 0 = 0 twice.
  2. Only one independent equation is left: x − 2y + z = 3.
  3. Two parameters: let y = s and z = t. Then x = 3 + 2s − t.

Every point of the plane: (x, y, z) = (3 + 2s − t, s, t), s, t ∈ ℝ. The three planes coincide.

Just for fun

Elimination is like a reality TV show for variables. Each round, one variable is voted off, until only one is left standing. Then, in a surprise twist, the others are brought back for the finale (back-substitution).

Hands-on activity

Design your own system

You need: paper, a pencil and a calculator.

  1. Pick a secret point with small integer coordinates, for example (2, −1, 3).
  2. Write three planes through it: choose any coefficients and compute the constant. For example 1x + 1y + 1z = 2 − 1 + 3 = 4, then 2x − y + z = ?, then x + 3y − 2z = ?. Make sure no equation is a combination of the others: check that the triple product of the normals is not 0.
  3. Swap systems with a family member (or put yours away for a day) and solve by elimination.
  4. Now break it on purpose: replace the third equation with (1) + (2). Solve again. What do you get? Then change only the constant of that new third equation and solve again.

What to look for: the first system returns your secret point. With (3) = (1) + (2) you get 0 = 0 and a line through the secret point. Changing only the constant gives 0 = k with k ≠ 0: no solution, a triangular prism.

Your turn

1

Solve x + y + z = 3, x − y + z = 1, x + y − z = 1.

2

Solve 2x + y − z = 1, x − y + 2z = 5, 3x + 2y + z = 10.

3

Solve x + 2y − z = 3, 2x + 4y − 2z = 5, x − y + z = 0, and describe the configuration.

4

Solve x + y = 3, y + z = 5, x + z = 4.

5

During elimination you get 0 = 0 exactly once, and the other equations are not multiples of each other. What is the intersection?

6

Solve x − y + z = 1, 2x + y − z = 2, x + 2y − 2z = 1.

7

Find where the planes x = 2, y = x + 1 and z = x + y meet.

More practice: warm-up

1

Solve x = 1, y = 2, x + y + z = 6.

2

Elimination ends with 0 = 4. What does that mean?

3

Elimination ends with 0 = 0 once and one free variable. What is the intersection?

4

Solve x + y = 5, y = 2, z = x.

5

Does (1, −1, 2) satisfy x + y + z = 2, 2x − y = 3 and z = 2?

More practice: on your own

1

Solve 3x − y + 2z = 7, x + y + z = 6, 2x + 3y − z = 5.

2

Solve x + y + z = 1, 2x + y + 3z = 4, x + 2z = 3.

3

Solve x + y + z = 1, 2x + y + 3z = 4, x + 2z = 5. Describe the configuration.

4

For which k does x + y + z = 2, x − y + z = 0, 2x + kz = 2 have infinitely many solutions?

5

A system's solution is x = 2 − t, y = 1, z = t. Describe it geometrically.

Challenge

1

Solve 2x + 3y − z = 4, x − 2y + 2z = −3, 3x + y + z = 2, and interpret the result.

2

For which values of a and b does the system x + y + z = 1, x + 2y + 3z = 2, x + 3y + az = b have (a) one solution, (b) no solution, (c) infinitely many solutions?

3

The planes x + y + z = 6, x − y + 2z = 5 and ax + y + z = b meet in a line. Find a and b, and describe the configuration.

Check your understanding

Quiz

1.

What is the solution of x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2?

2.

Elimination on a system of three planes ends with 0 = 5. Which statement is true?

3.

Elimination produces 0 = 0 once, and no two planes are parallel. The planes:

4.

Solve x + y = 3, y + z = 5, x + z = 4.

5.

The system x − 2y + z = 3, 2x − 4y + 2z = 6, −x + 2y − z = −3 has solutions:

6.

Elimination gives 0 = 2, and none of the three normals are parallel to each other. The configuration is:

For parents and teachers

Big idea (C4.4): the intersection of three planes is the solution of a system of three equations. Elimination finds it, and the ending tells the story: one point, 0 = 0 (a line or a plane), or 0 = k (no solution).

How to help: Have your teen label equations and write down each combination they use, for example "(1) + (3)". Insist on checking the final answer in all three original equations; it takes a minute and catches most errors.

Common difficulties:

  • Eliminating different variables in the two combinations, so a 2-by-2 system still has three unknowns.
  • Sign errors when subtracting equations. Writing the subtraction as adding the negative helps.
  • Treating 0 = 0 as "no solution". It means a redundant equation and infinitely many solutions.
  • Stopping at "no solution" without describing the configuration. Check whether any two normals are parallel.