When Different Equations Describe One Plane
A plane in \(\mathbb{R}^3\) can be written in many different ways. For instance, multiplying every term in an equation by the same nonzero number changes its appearance but not its solutions. When a system contains equations for the same plane, its solutions are all the points on that plane—not just one point or one line. A useful description of those points assigns arbitrary values to two coordinates and calculates the third.
Write a point in \(\mathbb{R}^3\) as the column \(\mathbf{x}=(x,y,z)^{\top}\). A plane equation has the form \(ax+by+cz=d\), where \(a,b,c,d\in\mathbb{R}\) and at least one of \(a,b,c\) is nonzero. The vector \(\mathbf{n}=(a,b,c)^{\top}\) is called a normal vector to the plane: its entries are the coefficients of the variables. The condition that the normal vector is nonzero ensures that the equation describes a plane rather than a different kind of solution set.
One immediate way to make a coincident equation is to multiply both sides by a nonzero scalar. For example, \(x+2y-z=3\) and \(-2x-4y+2z=-6\) have the same solutions: the second equation is the first multiplied by \(-2\). The scalar must be nonzero, and it must multiply the right-hand side as well as all three variable coefficients. The next result shows that this scaling test characterizes coincident planes completely.
Proof. If the four coefficients of the second equation are \(k\) times those of the first, where \(k\ne0\), then the second left-hand side equals \(k\) times the first left-hand side, and its right-hand side is \(kd\). Dividing the second equation by \(k\) recovers the first, so the equations have the same solutions.
For the reverse direction, suppose the two equations describe the same plane. Set \(\mathbf{n}=(a,b,c)^{\top}\), \(\mathbf{m}=(a',b',c')^{\top}\), and write the equations as \(\mathbf{n}\cdot\mathbf{x}=d\) and \(\mathbf{m}\cdot\mathbf{x}=d'\), using the usual dot product. Let \(P\) be their common plane. If \(\mathbf{u}\) satisfies \(\mathbf{n}\cdot\mathbf{u}=0\), take any \(\mathbf{p}\in P\). Then \(\mathbf{p}+\mathbf{u}\in P\), since \(\mathbf{n}\cdot(\mathbf{p}+\mathbf{u})=d+0=d\). Both points also satisfy the second equation, so subtracting their equations gives \(\mathbf{m}\cdot\mathbf{u}=0\). Reversing the roles of the equations shows that \(\mathbf{m}\cdot\mathbf{u}=0\) implies \(\mathbf{n}\cdot\mathbf{u}=0\). Thus the two normal vectors have the same vectors orthogonal to them.
Because \(\mathbf{n}\ne\mathbf{0}\), the number \(\mathbf{n}\cdot\mathbf{n}\) is positive. Define \(\mathbf{v}=\mathbf{n}/(\mathbf{n}\cdot\mathbf{n})\), so \(\mathbf{n}\cdot\mathbf{v}=1\). For any \(\mathbf{x}\in\mathbb{R}^3\), the vector \(\mathbf{x}-(\mathbf{n}\cdot\mathbf{x})\mathbf{v}\) is orthogonal to \(\mathbf{n}\). It is therefore orthogonal to \(\mathbf{m}\), by the equality of the sets of orthogonal vectors just proved. Consequently,
This holds for every \(\mathbf{x}\). Set \(k=\mathbf{m}\cdot\mathbf{v}\). The displayed equality says that \(\mathbf{m}=k\mathbf{n}\): taking \(\mathbf{x}=\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\) shows equality of the three coordinates. Since \(\mathbf{m}\ne\mathbf{0}\), \(k\ne0\). Finally, choose any \(\mathbf{p}\in P\). Then \(d'=\mathbf{m}\cdot\mathbf{p}=k(\mathbf{n}\cdot\mathbf{p})=kd\). All four coefficients have the required common factor. \(\square\)
Writing All the Solution Triples
If a system consists of equations for the same plane, a triple solves the system exactly when it satisfies one of the equations. To describe all such triples, choose a variable whose coefficient is nonzero, assign arbitrary real values to the other two variables, and solve for the chosen variable. The parameter values can be chosen independently; neither one is restricted by the other.
Proof. Choose any \(s,t\in\mathbb{R}\), and set \(x=s\), \(y=t\), and \(z=(d-as-bt)/c\). Since \(c\ne0\), this value is defined, and substitution gives \(ax+by+cz=as+bt+c(d-as-bt)/c=d\). Thus every pair produces a solution. Conversely, if \((x,y,z)\) is any solution, then \(cz=d-ax-by\), so \(z=(d-ax-by)/c\). Taking \(s=x\) and \(t=y\) produces that triple from the formula. Finally, the first two coordinates of a produced triple are \(s,t\), so two different parameter pairs cannot produce the same triple. \(\square\)
The condition \(c\ne0\) is only needed for solving specifically for \(z\). If \(c=0\), at least one of \(a,b\) is nonzero, so one can instead solve for \(x\) or \(y\), assigning the other two coordinates freely. The geometric idea is the same: one nonzero-coefficient equation imposes one constraint on three coordinates, leaving two coordinates arbitrary.
Worked Example: Three Scaled Equations for One Plane
Describe all solution triples of the system
Each equation is a multiple of the first, including its right-hand side: the second uses the factor \(2\), and the third uses the factor \(-3\). The theorem therefore shows that all three equations describe the same plane. To list its triples, set \(y=s\) and \(z=t\), where \(s,t\) are arbitrary real numbers. The first equation then gives \(x=4-s-t\). The family is
Check: Substitution into the first equation gives \((4-s-t)+s+t=4\). The left-hand side of the second is \(2(4-s-t)+2s+2t=8\), and the left-hand side of the third is \(-3(4-s-t)-3s-3t=-12\). Thus every displayed triple satisfies all three equations. Conversely, every solution must satisfy the first equation, and choosing \(s=y\), \(t=z\) forces \(x=4-s-t\), so none are omitted.
Worked Example: Solving for a Different Coordinate
Find a two-parameter description of the solutions to
The second equation is \(-3\) times the first, including the right-hand side, so the system describes one plane. Assign \(x=s\) and \(z=t\). Rearranging the first equation gives \(-y=5-2s-3t\), hence \(y=2s+3t-5\). The solution triples are
Check: In the first equation, the left-hand side becomes \(2s-(2s+3t-5)+3t=2s-2s-3t+5+3t=5\). In the second, it becomes \(-6s+3(2s+3t-5)-9t=-6s+6s+9t-15-9t=-15\). Thus every parameter pair gives a solution of both equations. Any solution has \(s=x\) and \(t=z\), and the first equation then forces \(y=2s+3t-5\), so the formula gives every solution.
Worked Example: Choosing \(x\) and \(y\) Freely
Describe the solutions to the three equations
The second and third equations are respectively \(-2\) and \(3\) times the first, on both sides. They all describe the same plane. Here the coefficient of \(z\) is \(-1\), so solve for \(z\): \(z=3x+2y-7\). Set \(x=s\) and \(y=t\) to obtain
Check: In the first equation, \(3s+2t-(3s+2t-7)=7\). Multiplying this equality by \(-2\) gives the second equation, with both sides \(-14\); multiplying it by \(3\) gives the third equation, with both sides \(21\). Conversely, a solution of the system satisfies the first equation, which determines \(z\) from its \(x\) and \(y\) values. Setting \(s=x\) and \(t=y\) therefore recovers every solution triple.
A Common Mistake: Comparing Only the Variable Coefficients
Proportional variable coefficients do not by themselves prove that two equations describe the same plane. The right-hand sides must have the same proportionality factor. For example, consider \(2x+y-z=3\) and \(4x+2y-2z=8\). The second set of variable coefficients is twice the first, but twice the first right-hand side would be \(6\), not \(8\). The equations do not pass the coincident-plane test.
A concrete substitution confirms the difference: \((x,y,z)=(1,1,0)\) satisfies the first equation because \(2(1)+1-0=3\), but it does not satisfy the second because \(4(1)+2(1)-2(0)=6\ne8\). A point on one plane but not the other rules out the claim that the planes are the same. The safe check is to compare the full coefficient quadruples \((a,b,c,d)\), not just the normal vectors \((a,b,c)\).
Reading the Family Geometrically
A parameterized formula such as \((4-s-t,s,t)\) is not a list of a few example points. It represents every point obtained as \(s\) and \(t\) range independently over \(\mathbb{R}\). For instance, \(s=0,t=0\) gives \((4,0,0)\), while \(s=1,t=2\) gives \((1,1,2)\); both lie on \(x+y+z=4\). Substitution verifies them: \(4+0+0=4\) and \(1+1+2=4\).
Coincident equations may make a system look as though it imposes several constraints, but exact proportionality shows that they all express the same constraint. Once one equation is retained, the two-parameter description makes the full set of solutions explicit. This representation is useful because it distinguishes a single point from an entire plane of solutions and gives a direct way to generate or check any solution triple.
Check Your Understanding
For each question, state the reasoning and check any proposed parameterization by substitution.
- Do \(x-2y+z=5\) and \(3x-6y+3z=15\) describe the same plane? Identify the scaling factor.
- Write all solution triples of \(x+2y-z=4\) by choosing \(x\) and \(y\) as parameters.
- Are \(2x+y+z=3\) and \(-4x-2y-2z=-8\) coincident? Explain why comparing only the variable coefficients is insufficient.
- For \(3x-y+2z=6\), choose \(x=s\) and \(z=t\) and derive a formula for \(y\).
- Explain why every solution of a single plane equation with a nonzero coefficient of \(z\) appears in the two-parameter formula that solves for \(z\).
