Two planes share a line
Two planes that are not parallel cross each other, and where they cross they share a whole straight line, like the crease of an open book or the line where a wall meets the floor. They never share just one point: two flat sheets that meet keep meeting along a straight line in both directions.
Take the planes x + y + z = 6 and x + 2y + 3z = 14. Their normals are (1, 1, 1) and (1, 2, 3). Neither is a multiple of the other, so the planes are not parallel, and they meet in a line. Finding that line means finding its direction and one point on it.
Two planes seen edge-on, looking straight along the line they share. In this view each plane is a line, and the shared line is the single point where they cross.
The direction: square to both normals
The line lies in the first plane, so it is perpendicular to that plane’s normal. It also lies in the second plane, so it is perpendicular to the second normal too. A direction at right angles to two vectors is what the vector product gives.
So the direction is (1, 1, 1) × (1, 2, 3) = (1 × 3 − 1 × 2, 1 × 1 − 1 × 3, 1 × 2 − 1 × 1) = (1, −2, 1).
Check with dot products: (1, −2, 1) · (1, 1, 1) = 1 − 2 + 1 = 0 and (1, −2, 1) · (1, 2, 3) = 1 − 4 + 3 = 0.
a × b stands perpendicular to both, as long as the parallelogram's area |a||b| sin θ
Swing b until it is parallel to a
Read a and b as the two normals. Their vector product a × b stands at right angles to both, and a direction at right angles to a plane’s normal runs along that plane, so a × b runs along both planes: it is the direction of their shared line. Swing b onto the line of a: the normals become parallel, a × b becomes zero, and there is no line.
A point: fix one coordinate
The line has direction (1, −2, 1), whose z-component is not 0, so it climbs through every height, including z = 0. Find where it passes through the floor by setting z = 0 in both equations.
That leaves x + y = 6 and x + 2y = 14, two equations in two unknowns. Subtract the first from the second: y = 8. Then x = 6 − 8 = −2. The point is (−2, 8, 0).
Check it in both planes: −2 + 8 + 0 = 6, and −2 + 2 × 8 + 3 × 0 = 14.
The floor z = 0. The first plane meets it along x + y = 6 and the second along x + 2y = 14. These two traces cross at (−2, 8), which is where the shared line passes through the floor.
The equation of the line
A point and a direction make a line: .
Check a second point. gives (0, 4, 2), and 0 + 4 + 2 = 6 and 0 + 2 × 4 + 3 × 2 = 14, so it is on both planes too.
In Cartesian form the line is .
Fixing a different coordinate
Any coordinate can be fixed. Setting x = 0 instead gives y + z = 6 and 2y + 3z = 14. Then y = 6 − z, so 2(6 − z) + 3z = 14, which is 12 + z = 14, so z = 2 and y = 4. The point (0, 4, 2) is the one found above at .
Setting z = 0 fails when the line never reaches the floor. The planes z = 3 and x + y = 4 have normals (0, 0, 1) and (1, 1, 0), and (0, 0, 1) × (1, 1, 0) = (−1, 1, 0), a level direction. Putting z = 0 into z = 3 gives 0 = 3, which is impossible: the line stays at height 3. Fix x = 0 instead: y = 4 and z = 3, the point (0, 4, 3).
The same line by elimination
The line can also be found without the vector product. Let , any number. Then and . Subtracting, , and then .
So the points of the line are , the same line as before.
Parallel planes
If the normals are multiples of each other, the planes face the same way, and their vector product is (0, 0, 0): there is no direction for a line. Such planes are either the same plane written twice or parallel planes that never meet.
The planes x + y + z = 6 and 2x + 2y + 2z = 5 have normals (1, 1, 1) and (2, 2, 2) = 2(1, 1, 1). Doubling the first equation gives 2x + 2y + 2z = 12, not 5, so the second plane is a different, parallel plane, and the two share no point at all. If the second had been 2x + 2y + 2z = 12, it would be the first plane written again, sharing every point.
The floor z = 0 again, for x + y + z = 6 and 2x + 2y + 2z = 5. Their traces, x + y = 6 and 2x + 2y = 5, are parallel and never cross, and the same is true at every height.
The usual mistakes
Expecting a single point. Two planes that meet share a line; it takes a third plane to cut that line at one point.
Adding the normals. (1, 1, 1) + (1, 2, 3) = (2, 3, 4) points out of the planes: (2, 3, 4) · (1, 1, 1) = 9, not 0.
Using one normal as the direction. A normal points straight out of its own plane, so it cannot lie along the plane.
Taking a point that is on only one plane. (6, 0, 0) satisfies x + y + z = 6 but gives 6 in the second equation, not 14. With z = 0, both equations have to hold at once.
The hip of a roof
In the application below, two faces of a hipped roof are planes, and the sloping edge where they meet, the hip, is their shared line. The vector product of the normals gives its direction, the peak is a point on both faces, and the eaves give where the hip ends.
Worked example: The Hip Where Two Faces of a Hipped Roof Meet
Question A hipped roof rises to a peak at P(0, 0, 5), in meters, with z up. Two of its faces lie in the planes x + 2z = 10 and y + 2z = 10, and they meet along a sloping edge called the hip, which runs down from P to the eaves, 3 m above the ground. (a) Find the Cartesian equation of the hip. (b) How long is the hip rafter, from P down to the eaves?
1.The normals are n1 = 102 and n2 = 012. The hip lies in both faces, so its direction is perpendicular to both of them.
The faces have normals n1 = 102 and n2 = 012, standing off the two shaded faces. 2.Take the vector product. The first component is 0 × 2 − 2 × 1 = −2, the second is 2 × 0 − 1 × 2 = −2, and the third is 1 × 1 − 0 × 0 = 1. So n1 × n2 = −2−21, and going down the roof the direction is 22−1.
The hip lies in both faces, so it is along n1 × n2 = −2−21. 3.P is on both faces, since 0 + 2 × 5 = 10 in each equation. So the hip is r = 005 + λ22−1.
P is on both faces, so the hip is r = 005 + λ22−1, going down the roof. The board writes s for λ. 4.(a) The coordinates are x = 2λ, y = 2λ and z = 5 − λ. Make λ the subject of each and set the three equal: x2 = y2 = z − 5−1.
(a) x = 2λ, y = 2λ, z = 5 − λ, so x2 = y2 = z − 5−1. 5.The eaves are at z = 3, so 5 − λ = 3 and λ = 2. The hip ends at (4, 4, 3). Check: 4 + 2 × 3 = 10 in both equations.
The eaves are at z = 3: λ = 2, and the hip ends at (4, 4, 3). 6.(b) The rafter runs from (0, 0, 5) to (4, 4, 3), so its length is √42 + 42 + (−2)2 = √36 = 6 m. Check: it is 2 steps of the direction vector, whose length is √4 + 4 + 1 = 3, and 2 × 3 = 6.
(b) The hip rafter is √16 + 16 + 4 = 6 m long.
Answer: (a) x2 = y2 = z − 5−1; (b) 6 m, from P down to the corner of the eaves at (4, 4, 3)
Common mistakes
- Taking one of the normals as the direction of the hip. A normal points out of its face, and the hip lies along the face, so the direction must be perpendicular to BOTH normals.
- Writing the Cartesian form as x2 = y2 = z − 51, dropping the minus sign. The height falls as the hip runs out from the peak, so the third component of the direction is −1.
More planes and the vector product problems, worked step by step →