When Three Planes Share a Line
A plane in \(\mathbb{R}^3\) is the set of points \((x,y,z)\) satisfying one linear equation \(ax+by+cz=d\), where \(a,b,c,d\) are real numbers and \(a,b,c\) are not all zero. A system of three plane equations asks for the points that lie on all three planes at once. Sometimes those common points are just one point; in another important situation, they form a whole line.
To describe a line in coordinates, we use a real parameter \(t\). For instance, the points \((1+t,t,2-2t)\), as \(t\) ranges over \(\mathbb{R}\), form a line: the coordinates change at a constant rate as \(t\) changes. A parameter description gives every point on the line, not just a few sample points. To show that three planes share this line, we must verify that every point in the description satisfies all three equations.
A useful way for three planes to share a line is for the third equation to be a consequence of the first two. In that case, every point satisfying the first two equations automatically satisfies the third. The following result makes this precise, provided the first two planes already have a line in common.
Proof. Write the first two equations as \(L_1(x,y,z)=d_1\) and \(L_2(x,y,z)=d_2\), where each \(L_i\) is a linear expression in \(x,y,z\). Suppose the third equation is \(\alpha L_1(x,y,z)+\beta L_2(x,y,z)=\alpha d_1+\beta d_2\). If a point satisfies the first two equations, then its left side in the third equation equals \(\alpha d_1+\beta d_2\), so it satisfies the third equation too. Thus every common solution of the first two is a common solution of all three. Conversely, any solution of all three satisfies the first two, because those are among the three equations. The common solution sets are therefore equal. By hypothesis, that set is a line. \(\square\)
Finding the Common Line by Substitution
Worked Example: A Third Plane Containing the Common Line
Find all common solutions of the three planes \(x+y+z=4\), \(x-y=0\), and \(3x-y+z=4\).
Start with the first two equations. The equation \(x-y=0\) gives \(y=x\). Let their common value be \(t\), so \(x=t\) and \(y=t\). Substituting into the first equation gives \(t+t+z=4\), hence \(z=4-2t\). The first two planes therefore meet in the parameterized line
Check this description in all three equations. In the first equation, \(t+t+(4-2t)=4\). In the second, \(t-t=0\). In the third, \(3t-t+(4-2t)=4\). Thus every point on the line satisfies all three equations.
We can also see why the third equation adds no new restriction: adding the first equation to twice the second gives \((x+y+z)+2(x-y)=4+2\cdot0\), which simplifies to \(3x-y+z=4\). Conversely, a solution of all three must satisfy the first two, whose solutions have already been described. So the displayed line is exactly the common solution set.
The theorem explains the role of a linear combination: it makes one equation follow from the others. Geometry adds an important condition. If the first two planes meet in a line, then a third plane given by their linear combination contains that whole line. The third plane need not be parallel to either of the first two; it can be a different plane passing through the same line.
When Does a Third Plane Contain the Whole Line?
In the example below, the first two planes have a particularly convenient parameterization. We will use them to prove a converse as well: for this pair, any plane that contains their entire common line has an equation that is a linear combination of the first two equations. The concrete form of the pair lets us show every step directly.
Proof. Solving the first two equations gives \(x=y+1\). Set \(y=t\); then \(x=1+t\), and the first equation gives \(z=2-2t\). Their common line is therefore \((x,y,z)=(1+t,t,2-2t)\) for \(t\in\mathbb{R}\). First suppose the third plane contains this entire line. Substitution into \(Ax+By+Cz=D\) gives \((A+B-2C)t+(A+2C)=D\) for every real \(t\). Taking \(t=0\) gives \(D=A+2C\). Taking \(t=1\) and using that equality gives \(A+B-2C=0\), or \(B=2C-A\). Now multiply \(x+y+z=3\) by \(C\), multiply \(x-y=1\) by \(A-C\), and add. The resulting equation is \(Ax+(2C-A)y+Cz=A+2C\), which is \(Ax+By+Cz=D\) by the two identities just obtained. This proves that the third equation is a linear combination. Conversely, if its equation is a linear combination of the first two, every point satisfying those two equations satisfies the third, by substitution into the combination. So the third plane contains their entire common line. \(\square\)
This result also clarifies the distinct-plane case. In this particular pair, if the third plane is distinct from both of the first two, both multipliers in its linear combination must be nonzero. If one multiplier were zero, the equation would be a nonzero multiple of just one of the original plane equations and would describe that same plane.
Worked Example: Checking a Third Plane Against the Whole Line
Consider the planes \(x+y+z=3\), \(x-y=1\), and \(2y+z=2\). Determine their common solution set.
The first two equations give \(x=1+t\), \(y=t\), and \(z=2-2t\), as above. Substitution into the third equation gives \(2t+(2-2t)=2\), which is true for every real \(t\). Hence the whole line satisfies the third equation, and the common solution set is
For an independent check, the third equation is the first equation minus the second: \((x+y+z)-(x-y)=3-1\), which simplifies to \(2y+z=2\). Every solution of the first two must therefore satisfy the third. Conversely, a solution of all three satisfies the first two and so lies on the displayed line.
Building Three Planes Around a Line
A shared line can also be used to construct examples. Choose two plane equations, solve them using one free coordinate, and then form a third equation by combining the first two. The third plane will contain the line automatically. The next example carries out these steps and checks the result in each equation.
Worked Example: Constructing a Shared Line
Consider the planes \(x+y=3\), \(z-2x=-1\), and \(2y+z=5\). Find their common solutions.
From the first plane, \(y=3-x\). From the second, \(z=2x-1\). Let \(x=2+t\), where \(t\) can be any real number. Then \(y=3-(2+t)=1-t\) and \(z=2(2+t)-1=3+2t\). The candidate common line is
Substitute into the first equation: \((2+t)+(1-t)=3\). Substitute into the second: \((3+2t)-2(2+t)=3+2t-4-2t=-1\). Substitute into the third: \(2(1-t)+(3+2t)=2-2t+3+2t=5\). All three equations hold for every real \(t\).
The third equation is also the sum of twice the first equation and the second: \(2(x+y)+(z-2x)=2\cdot3+(-1)\), which simplifies to \(2y+z=5\). Thus it follows from the first two. Conversely, every common solution must satisfy the first two, and those equations give exactly the parameterized line above. This proves that no common solutions have been missed.
Common Pitfall: Checking Only One Point
A frequent error is to substitute one point from a proposed line into the third equation and conclude that the third plane contains the line. One point only establishes that the third plane contains that point. To show that it contains the entire line, the equation must hold for every parameter value.
For example, in the first worked example the common line of the first two planes is \((x,y,z)=(t,t,4-2t)\). The tempting but incorrect claim that \(x+z=4\) contains this line may seem plausible because at \(t=0\) the point is \((0,0,4)\), and \(x+z=0+4=4\). But at \(t=1\), the line gives \((1,1,2)\), and \(x+z=1+2=3\ne4\). So that plane contains the point for \(t=0\), not the whole line. In contrast, checking the proposed line in the valid third equation \(3x-y+z=4\) gives \(3t-t+(4-2t)=4\), which holds for every real \(t\).
There are two complementary ways to verify a shared-line claim: substitute the parameterized coordinates into every plane equation, or show that the third equation follows by adding multiples of the first two. For a complete solution, also explain why every common solution appears in the parameterization; solving the first two equations and assigning their remaining coordinate the parameter does this.
Check Your Understanding
For each question, use substitution to verify any proposed parameter description.
- Find the common solution set of \(x+y+z=5\), \(x-y=1\), and \(2x+z=6\).
- Show that \(x+y+z=2\), \(x-y=0\), and \(3x-y+z=2\) share a line. Give all common solutions using a parameter.
- For the line \((x,y,z)=(1+t,t,2-2t)\), test whether \(x+z=3\) contains the whole line. If not, find a value of \(t\) that disproves the claim.
- Explain why a third equation formed by adding twice one plane equation to another is satisfied by every point on the common line of those two planes.
- For the planes \(x+y+z=3\) and \(x-y=1\), substitute \((1+t,t,2-2t)\) into \(Ax+By+Cz=D\). What conditions on \(A,B,C,D\) make the equation true for every real \(t\)?
