How Two Planes Can Meet
Two planes in three-dimensional space can have three different relationships: they can be parallel and distinct, coincide, or intersect in a line. To determine which situation applies, compare the planes’ normal vectors and, when they intersect, solve their equations together.
A plane in general form is written \(ax+by+cz=d\), where \(a,b,c\) are not all zero. Its normal vector is \(\mathbf{n}=\langle a,b,c\rangle\). The normal vector is perpendicular to every direction that lies in the plane.
The cross product \(\mathbf{n}_1\times\mathbf{n}_2\) gives an immediate test. If it is nonzero, the normals are not parallel, so the planes intersect in a line. If it is zero, the normals are parallel, and the constants determine whether the planes coincide or are distinct.
Classifying the Planes
Suppose one normal vector is a nonzero scalar multiple of the other: \(\mathbf{n}_2=\lambda\mathbf{n}_1\), where \(\lambda\ne 0\). The planes have the same orientation. If also \(d_2=\lambda d_1\), the second equation is exactly \(\lambda\) times the first, so they describe the same plane. If \(d_2\ne\lambda d_1\), they are distinct parallel planes and have no common point.
When the normals are not proportional, the planes meet in a line. Its direction must lie in both planes, so it must be perpendicular to both normals. The cross product \(\mathbf{n}_1\times\mathbf{n}_2\) supplies such a direction. To complete the parametrization, find one point satisfying both plane equations.
| Relationship | Test | Intersection |
|---|---|---|
| Intersecting planes | \(\mathbf{n}_1\times\mathbf{n}_2\ne\mathbf{0}\) | A line |
| Coincident planes | \(\mathbf{n}_2=\lambda\mathbf{n}_1\) and \(d_2=\lambda d_1\) for some \(\lambda\ne0\) | The whole plane |
| Distinct parallel planes | \(\mathbf{n}_2=\lambda\mathbf{n}_1\) but \(d_2\ne\lambda d_1\) | No points |
These tests assume each equation represents a plane, so its three coefficients of \(x,y,z\) are not all zero. The scalar \(\lambda\) must be nonzero because a plane has a nonzero normal vector.
A Method for Finding the Intersection Line
When the normals are not proportional, there are two tasks: find a point on both planes and find a direction vector parallel to both planes. Often the quickest way to find a point is to set one coordinate to zero and solve the resulting pair of equations. If that choice does not yield a solution, choose another coordinate or a convenient value instead.
From each plane equation, take the coefficients of \(x,y,z\) in order.
Compute \(\mathbf{n}_1\times\mathbf{n}_2\). If it is zero, compare the constants using the same scale factor as the normals. If it is nonzero, the planes intersect in a line.
Solve the two plane equations together, choosing a convenient value for one variable when possible.
Use the point and the cross product direction in a parametric equation, then substitute the parametrization into both planes.
If \(\mathbf{r}_0=\langle x_0,y_0,z_0\rangle\) is a common point and \(\mathbf{v}=\mathbf{n}_1\times\mathbf{n}_2\), the line is \(\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{v}\), where \(t\in\mathbb{R}\). Any nonzero scalar multiple of \(\mathbf{v}\) gives the same line; a simpler direction vector may therefore be used.
Worked Intersection
Worked Example: Find the Line of Intersection
Find the intersection of \(x+y+z=4\) and \(x-y+2z=2\).
The normal vectors are \(\mathbf{n}_1=\langle1,1,1\rangle\) and \(\mathbf{n}_2=\langle1,-1,2\rangle\). Their cross product is
This direction is nonzero, so the planes intersect in a line. To find a point, set \(z=0\). The equations become \(x+y=4\) and \(x-y=2\). Adding gives \(2x=6\), so \(x=3\). Substituting into \(x+y=4\) gives \(3+y=4\), hence \(y=1\). Thus \(\mathbf{r}_0=\langle3,1,0\rangle\) lies on both planes.
The line is therefore
In component form, \(x=3+3t\), \(y=1-t\), and \(z=-2t\). Substitution verifies the answer: in the first plane, \((3+3t)+(1-t)+(-2t)=4\); in the second, \((3+3t)-(1-t)+2(-2t)=2\). Both equations hold for every real \(t\).
Why the Cross Product Gives the Direction
For any point on the intersection line, a direction vector from one point on that line to another lies within both planes. A direction in the first plane is perpendicular to \(\mathbf{n}_1\), and a direction in the second is perpendicular to \(\mathbf{n}_2\). The cross product of two nonparallel vectors is perpendicular to each of them, so \(\mathbf{n}_1\times\mathbf{n}_2\) is a direction that lies parallel to both planes.
To see why this describes the entire intersection, write the planes as \(\mathbf{n}_1\cdot\mathbf{r}=d_1\) and \(\mathbf{n}_2\cdot\mathbf{r}=d_2\). The cross product is perpendicular to both normals, so adding \(t(\mathbf{n}_1\times\mathbf{n}_2)\) to a common point leaves both dot products unchanged. Thus every point on the parametrized line lies in both planes.
Conversely, if \(\mathbf{r}\) is any other common point, then \(\mathbf{w}=\mathbf{r}-\mathbf{r}_0\) satisfies \(\mathbf{n}_1\cdot\mathbf{w}=0\) and \(\mathbf{n}_2\cdot\mathbf{w}=0\). When the normals are not parallel, the vectors perpendicular to both normals form a one-dimensional direction, spanned by \(\mathbf{n}_1\times\mathbf{n}_2\). Therefore \(\mathbf{w}=t(\mathbf{n}_1\times\mathbf{n}_2)\) for some real \(t\), proving that every common point is on the stated line.
A common point exists in this case: since the normals are not proportional, the two equations impose two independent linear constraints on three coordinates. One can set a remaining coordinate to a suitable value and solve for the other two. Hence the intersection is not empty and, by the argument above, is exactly a line.
Coincident and Parallel Planes
Worked Example: Identify Coincident Planes
Classify \(2x-4y+6z=8\) and \(x-2y+3z=4\).
The first normal is \(\langle2,-4,6\rangle\), and the second is \(\langle1,-2,3\rangle\). The second normal is one-half the first. The right-hand side is also scaled by one-half: \(8/2=4\). In fact, dividing every term of the first equation by \(2\) gives \(x-2y+3z=4\), exactly the second equation. The planes coincide.
For a direct check, the point \(\langle4,0,0\rangle\) satisfies the first equation because \(2(4)-4(0)+6(0)=8\), and it satisfies the second because \(4-2(0)+3(0)=4\). The equations describe the same plane, not just a single shared point.
Worked Example: Identify Distinct Parallel Planes
Classify \(2x-y+3z=5\) and \(4x-2y+6z=13\).
The second normal vector, \(\langle4,-2,6\rangle\), is twice the first normal vector, \(\langle2,-1,3\rangle\). If the equations represented the same plane, the second right-hand side would also be twice the first: \(2(5)=10\). But its right-hand side is \(13\), not \(10\).
Equivalently, doubling the first equation gives \(4x-2y+6z=10\). A point common to both planes would then have to satisfy both \(4x-2y+6z=10\) and \(4x-2y+6z=13\), which would require \(10=13\). That is impossible. The planes are distinct and parallel, so there is no intersection.
A Common Mistake
A direction vector for the intersection line must be perpendicular to both normal vectors. It is not enough to choose a vector that merely combines their components in a plausible-looking way.
For the normals \(\langle1,1,1\rangle\) and \(\langle1,-1,2\rangle\) from the first example, an incorrect choice would be their sum, \(\langle2,0,3\rangle\). Its dot product with the first normal is \(2+0+3=5\), not zero, so it cannot be a direction lying in that plane. The correct cross product direction is \(\langle3,-1,-2\rangle\); its dot products with the two normals are \(3-1-2=0\) and \(3+1-4=0\). It is perpendicular to both, as required.
Check Your Understanding
Answer each question, showing the classification test or substitution that supports your conclusion.
- What relationship between two normal vectors shows that the planes are not parallel?
- Two normals are related by \(\mathbf{n}_2=3\mathbf{n}_1\), and the second constant is three times the first. What is the relationship between the planes?
- If the normals are proportional but the constants do not have the same scale factor, how many common points do the planes have?
- For intersecting planes, what vector gives a direction for their intersection line?
- Why should a proposed direction vector have dot product zero with each plane’s normal?