The Geometry of Three Planes

A point, a line, or nowhere at all.

Each equation is a plane

A linear equation in two unknowns, such as x + y = 3, is a straight line in the plane. A linear equation in three unknowns, such as x + y + z = 3, is a flat plane in space: it holds at (3, 0, 0), at (0, 3, 0), at (0, 0, 3) and at (1, 1, 1), and at every other point of the flat sheet through them.

Three such equations are three planes, and a solution of all three is a point that lies on all three planes at once.

How three planes can meet

Three planes in space can meet at exactly one point, like two walls and a floor meeting at a corner. They can share a whole line, like the pages of an open book meeting at its spine. Or they can have no point in common at all.

There are two ways to have no common point. Two of the planes can be parallel, so they never meet. Or no two are parallel, each pair meets along a line, and the three lines are parallel, like the three long edges of a triangular prism. (If all three equations describe the same plane, every point of that plane is a solution.)

A non-zero determinant: one point

Write the three equations as A X = b. If det A is not 0, A has an inverse, and X = A⁻¹ b is one column and only one. So the three planes meet at exactly one point.

For A = (1 2 3; 0 1 4; 5 6 0), det A = 1. With the totals 14, 14 and 17, the inverse gives x = 1, y = 2 and z = 3, so the planes x + 2y + 3z = 14, y + 4z = 14 and 5x + 6y = 17 meet at the single point (1, 2, 3).

A zero determinant

Now take the planes x + y + z = 3, 2x + y − z = 2 and 3x + 2y = 5. Their matrix is B = (1 1 1; 2 1 −1; 3 2 0). Row 3, which is 3, 2, 0, is row 1 plus row 2: 1 + 2 = 3, 1 + 1 = 2 and 1 + (−1) = 0.

Expanding along row 1, the three minors are 1 × 0 − (−1) × 2 = 2, then 2 × 0 − (−1) × 3 = 3, then 2 × 2 − 1 × 3 = 1. So det B = 1 × 2 − 1 × 3 + 1 × 1 = 0. B has no inverse, so these planes cannot meet at exactly one point.

Totals that agree: a line

Add the first two equations: (x + y + z) + (2x + y − z) = 3 + 2, which is 3x + 2y = 5. That is the third equation exactly, totals and all. So every point on the first two planes is already on the third, and the third equation adds nothing new.

The first two planes meet along a line. Taking the first equation from the second gives x − 2z = −1, so x = 2z − 1, and then y = 3 − x − z = 4 − 3z. Writing z = t, the line is x = 2t − 1, y = 4 − 3t, z = t, and every point on it is a solution.

Check three of them. At t = 0 the point is (−1, 4, 0): −1 + 4 + 0 = 3, −2 + 4 − 0 = 2 and −3 + 8 = 5. At t = 1 it is (1, 1, 1): 1 + 1 + 1 = 3, 2 + 1 − 1 = 2 and 3 + 2 = 5. At t = 2 it is (3, −2, 2): 3 − 2 + 2 = 3, 6 − 2 − 2 = 2 and 9 − 4 = 5. The three planes share this whole line.

Totals that disagree: nothing

Change the third total from 5 to 9, so the third plane is 3x + 2y = 9. Every point on the first two planes still has 3x + 2y = 3 + 2 = 5, and 5 is not 9. So no point lies on all three planes, and the system has no solution.

No two of these planes are parallel, because no row of B is a multiple of another. So each pair of planes meets along a line. The vector product of two normals gives the direction of their line: (1, 1, 1) × (3, 2, 0) = (−2, 3, −1) and (2, 1, −1) × (3, 2, 0) = (2, −3, 1), both along (2, −3, 1), the same direction as the line of the first two planes. The three lines are parallel, and the planes make the three faces of a triangular prism.

x + y + z = 32x + y − z = 23x + 2y = 93x + 2y + 0z3x + 2y + 4z

det = 0, but the third total is 9, not 3 + 2 = 5: no point satisfies all three equations, and each pair of planes meets along its own line, parallel to the other two

Make the three planes share a line

Each equation above the drawing is written in the color of its plane, and the gold plane is the third. At 3x + 2y = 9 the planes make a prism, each pair meeting along its own line. Drag the gold plane: its total changes, and only at 3 + 2 = 5 do the three lines close up into one. Then give the third equation a z term.

Giving the third equation a z term

Change the third plane to 3x + 2y + 4z = 9. Its row is now 3, 2, 4, which is no longer row 1 plus row 2, and expanding along row 1 gives 1 × (1 × 4 − (−1) × 2) − 1 × (2 × 4 − (−1) × 3) + 1 × (2 × 2 − 1 × 3) = 6 − 11 + 1 = −4.

The determinant is not 0, so there is exactly one common point. It lies on the line of the first two planes, x = 2t − 1, y = 4 − 3t, z = t, and putting that line into the third equation gives 3(2t − 1) + 2(4 − 3t) + 4t = 5 + 4t = 9, so t = 1 and the point is (1, 1, 1). With the total 13 instead, 5 + 4t = 13 gives t = 2 and the point (3, −2, 2). Whatever the total, there is one point, because the third plane cuts across the line instead of running beside it.

Parallel planes

A zero determinant can also come from two parallel planes. In x + y + z = 3, 2x + 2y + 2z = 10 and x − y = 0, row 2 is twice row 1, so the determinant is 0. Halving the second equation gives x + y + z = 5, which cannot hold together with x + y + z = 3: the first two planes are parallel and never meet, so there is no solution. Had the second total been 6, the first two equations would have been the same plane.

Deciding

First find the determinant of the matrix. If it is not 0, the three planes meet at exactly one point, and the inverse, or elimination, finds it.

If it is 0, there is not exactly one point, and the totals decide between the other cases. Combine the equations to remove the unknowns, as adding the first two did above. If the totals agree, as 3 + 2 = 5 did, there are infinitely many solutions, usually a whole line. If they contradict each other, as 5 and 9 did, there are none.

The usual mistakes

Reading a zero determinant as "no solution". With the totals 3, 2 and 5 the determinant is 0 and there are infinitely many solutions.

Expecting parallel planes whenever there is no solution. In the prism no two planes are parallel; each pair meets, but the three lines never meet.

Stopping at a point on two of the planes. (1, 1, 1) lies on x + y + z = 3 and on 2x + y − z = 2, but 3 × 1 + 2 × 1 = 5, not 9, so it is not on the plane 3x + 2y = 9. A solution has to satisfy every equation, so the check uses all three.

Treating the determinant as a coordinate or a distance. It only decides whether there is exactly one point.

Laser light and mixed boxes

In the first application below, three sheets of laser light at a concert are three planes whose determinant is 4, so they cross at a single point, where a mirror ball hangs. In the second, three weighings give a determinant of 0: with the weights as recorded the planes share a line, and with one weighing recorded differently they form a prism with no common point.

Worked example: Three Sheets of Laser Light at a Concert That Cross at a Single Point

Question At a concert, three flat sheets of laser light fill the planes x + z = 5, −y + 2z = 5 and 2x − y = 3, with coordinates in meters and z measured up from the stage. A mirror ball is to hang where all three sheets cross. (a) Find the determinant of the matrix of coefficients, and explain what it tells you about how the three planes meet. (b) Find the point where the mirror ball hangs.

  1. 1.The coefficients of x, y and z make the matrix 1010−122−10, with one row for each sheet.

    stagex + z = 5−y + 2z = 52x − y = 3x + z = 5−y + 2z = 52x − y = 3
    stagex + z = 5−y + 2z = 52x − y = 3x + z = 5−y + 2z = 52x − y = 3
    The three sheets are three planes: x + z = 5, −y + 2z = 5 and 2x − y = 3.

  2. 2.Expand along the first row. Its middle entry is 0, so only two 2 × 2 determinants are needed: 1 × ((−1) × 0 − 2 × (−1)) − 0 + 1 × (0 × (−1) − (−1) × 2) = 2 + 2 = 4.

    stagex + z = 5−y + 2z = 52x − y = 31010−122−10= 1 × 2 − 0 + 1 × 2 = 4
    stagex + z = 5−y + 2z = 52x − y = 31010−122−10= 1 × 2 − 0 + 1 × 2 = 4
    Expanding along the first row, whose middle entry is 0: 1010−122−10 = 2 + 2 = 4.

  3. 3.(a) The determinant is 4, not 0, so the matrix has an inverse and the three equations have exactly one solution: the three planes meet at a single point.

    stagex + z = 5−y + 2z = 52x − y = 34 is not 0: one point on all three
    stagex + z = 5−y + 2z = 52x − y = 34 is not 0: one point on all three
    (a) The determinant is not 0, so the three planes have exactly one point in common.

  4. 4.Elimination is quicker here than the inverse. The first sheet gives x = 5 − z, and then the third gives y = 2x − 3 = 2(5 − z) − 3 = 7 − 2z.

    stagex + z = 5−y + 2z = 52x − y = 3x = 5 − zy = 2x − 3 = 7 − 2z
    stagex + z = 5−y + 2z = 52x − y = 3x = 5 − zy = 2x − 3 = 7 − 2z
    The first sheet gives x = 5 − z, and then the third gives y = 2x − 3 = 7 − 2z.

  5. 5.Substitute both into the second: −(7 − 2z) + 2z = 5, so 4z − 7 = 5, 4z = 12 and z = 3. Then x = 5 − 3 = 2 and y = 7 − 2 × 3 = 1.

    stagex + z = 5−y + 2z = 52x − y = 3−(7 − 2z) + 2z = 5, so 4z = 12z = 3, x = 2, y = 1
    stagex + z = 5−y + 2z = 52x − y = 3−(7 − 2z) + 2z = 5, so 4z = 12z = 3, x = 2, y = 1
    In the second: −(7 − 2z) + 2z = 5, so 4z = 12, z = 3, x = 2 and y = 1.

  6. 6.(b) The mirror ball hangs at (2, 1, 3), which is 3 m above the stage. Check in all three sheets: 2 + 3 = 5, −1 + 2 × 3 = 5 and 2 × 2 − 1 = 3.

    stagex + z = 5−y + 2z = 52x − y = 33 m(2, 1, 3)the ball hangs at (2, 1, 3)check: 2 + 3 = 5, −1 + 6 = 5, 4 − 1 = 3
    stagex + z = 5−y + 2z = 52x − y = 33 m(2, 1, 3)the ball hangs at (2, 1, 3)check: 2 + 3 = 5, −1 + 6 = 5, 4 − 1 = 3
    (b) The mirror ball hangs at (2, 1, 3), 3 m above the stage, on all three sheets.

Answer: (a) The determinant is 4, which is not 0, so the three planes meet at exactly one point; (b) the mirror ball hangs at (2, 1, 3), 3 m above the stage

Common mistakes

  • Stopping at a point that satisfies two of the equations. Two planes meet along a whole line; only a point that satisfies all three equations is on all three sheets, so the check must use each one.
  • Reading the determinant 4 as a coordinate or a distance. The determinant decides only how the planes meet: not 0 means one point, and 0 means a line of common points or none at all.

More determinants and inverses problems, worked step by step →

Worked example: Three Weighings of Mixed Boxes: What They Fix, and a Weighing Recorded Differently

Question A warehouse stocks boxes of three kinds, weighing x, y and z kg each. Three loads are weighed: 1 box of the first kind, 1 of the second and 3 of the third weigh 13 kg; 3 boxes of the first kind weigh 6 kg; and 4 of the first kind, 1 of the second and 3 of the third weigh 19 kg. (a) Show that the matrix of the three equations has determinant 0. Find what the weighings do tell you, given that every box weighs more than nothing, and describe how the three planes meet. (b) Suppose the third load had been recorded as 13 kg. Show that the equations then have no solution, and describe how the three planes lie.

  1. 1.The loads give x + y + 3z = 13, 3x = 6 and 4x + y + 3z = 19, with the matrix 113300413.

    x + y + 3z = 133x = 64x + y + 3z = 19x + y + 3z = 133x = 64x + y + 3z = 19
    x + y + 3z = 133x = 64x + y + 3z = 19x + y + 3z = 133x = 64x + y + 3z = 19
    Each load is a plane, and all three contain the line where the first two meet.

  2. 2.Expand along the second row, which has two zeros. Its first entry takes the sign −, so the determinant is −3 × 1313 + 0 − 0 = −3 × (3 − 3) = 0. The matrix has no inverse, and there is no single solution.

    x + y + 3z = 133x = 64x + y + 3z = 19113300413= −3 × (3 − 3) = 0expanded along row 2, with the sign −
    x + y + 3z = 133x = 64x + y + 3z = 19113300413= −3 × (3 − 3) = 0expanded along row 2, with the sign −
    Expanding along the second row, which has two zeros: 113300413 = −3 × (3 − 3) = 0.

  3. 3.Row 3 is row 1 plus row 2, because the third load is the first two loads together. Its weight agrees, as 13 + 6 = 19, so the third equation adds nothing new.

    x + y + 3z = 133x = 64x + y + 3z = 19row 3 = row 1 + row 213 + 6 = 19: the totals agree
    x + y + 3z = 133x = 64x + y + 3z = 19row 3 = row 1 + row 213 + 6 = 19: the totals agree
    Row 3 is row 1 plus row 2, and 13 + 6 = 19: the third equation repeats the first two.

  4. 4.The second load gives x = 2, and then the first gives 2 + y + 3z = 13, so y + 3z = 11. The second and third kinds always go on the scale as 1 box and 3 boxes together, so only y + 3z can be found.

    x + y + 3z = 133x = 64x + y + 3z = 193x = 6, so x = 22 + y + 3z = 13, so y + 3z = 11
    x + y + 3z = 133x = 64x + y + 3z = 193x = 6, so x = 22 + y + 3z = 13, so y + 3z = 11
    The second load gives x = 2, and the first then gives y + 3z = 11.

  5. 5.(a) Each box of the first kind weighs 2 kg, but y and z cannot be separated: y = 5 with z = 2 fits, and so does y = 2 with z = 3. The three planes meet along the line x = 2, y + 3z = 11.

    x + y + 3z = 133x = 64x + y + 3z = 19x = 2, y + 3z = 11x = 2 kg, but only y + 3z = 11 is knowny = 5, z = 2 or y = 2, z = 3
    x + y + 3z = 133x = 64x + y + 3z = 19x = 2, y + 3z = 11x = 2 kg, but only y + 3z = 11 is knowny = 5, z = 2 or y = 2, z = 3
    (a) The three planes meet along the line x = 2, y + 3z = 11: each box of the first kind weighs 2 kg, and y and z cannot be separated.

  6. 6.(b) With 13 kg, the third load, which is the first two together, would weigh both 13 + 6 = 19 kg and 13 kg, so no weights fit all three equations. No two of the planes are parallel, so each pair meets along a line, but the three lines are parallel: the planes form a triangular prism, with no point on all three. One of the weighings must be wrong.

    x + y + 3z = 133x = 64x + y + 3z = 13the same boxes: 19 kg and 13 kgthree parallel lines: no common point
    x + y + 3z = 133x = 64x + y + 3z = 13the same boxes: 19 kg and 13 kgthree parallel lines: no common point
    (b) With 13 kg, each pair of planes meets along a line, but the three lines are parallel: a triangular prism with no point on all three.

Answer: (a) The determinant is 0, because the third load is the first two together; each box of the first kind weighs 2 kg, but only y + 3z = 11 is known, and the three planes meet along the line x = 2, y + 3z = 11; (b) the third load would weigh both 19 kg and 13 kg, so there is no solution: the planes form a triangular prism with no common point

Common mistakes

  • Deciding from the zero determinant alone that there is no solution. A zero determinant means there is not exactly one solution; whether there is a line of solutions or none depends on whether the totals agree, as 13 + 6 = 19 does here.
  • Expecting three planes with no common point to include two parallel planes. No row of this matrix is a multiple of another, so no two planes are parallel; they cross in pairs along three parallel lines, like the three faces of a triangular prism.

More determinants and inverses problems, worked step by step →

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