CP 3.8 - How three planes intersect

You already know

For $\mathbf{A}\mathbf{x}=\mathbf{b}$, a non-zero determinant gives a unique solution $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$. A zero determinant means the inverse method is unavailable; it does not decide whether the equations are consistent.

Consistent and inconsistent systems

A system is consistent if at least one point satisfies every equation, and inconsistent if no point does. Three planes may have one common point, a common line, or a whole common plane if all three coincide. An inconsistent system has no common point even if every pair intersects.

Key idea

The determinant separates “one solution” from “not one solution”. If it is zero, the constants still decide between infinitely many solutions and no solution.

Sheaf of planes

A sheaf is an arrangement of planes passing through the same line. For three distinct planes, each pair has that same intersection line. For example, $x=0$, $y=0$ and $x+y=0$ form a sheaf along the $z$-axis. Three planes with three different intersection lines do not form a sheaf.

Worked example 1: the third equation is a sum

Question. Describe the intersection of

and give all common points.

Plan. Test whether the third equation follows from the first two, then solve the two independent equations with one variable free.

Put $x=t$, where $t$ is any real number. Then $z=2-t$.

Why this step? Two independent equations leave one degree of freedom. Introducing $t$ describes the entire line rather than selecting one point from it.

Answer. The three distinct planes form a sheaf, with common line

Check. Substitution gives $t+1+(2-t)=3$, $t-1+(2-t)=1$, and $2t+2(2-t)=4$, for every real $t$.

Worked example 2: a weighted sum

Question. Describe the common intersection of

Plan. Compare the third equation with twice the first plus the second, then use the known intersection of the first two.

The first two equations still give $y=1$ and $x+z=2$.

Why compare entire equations? Matching only the variable coefficients would miss a possible contradiction in the constant.

Answer. The planes form a sheaf with common line $(x,y,z)=(t,1,2-t)$, $t\in\mathbb{R}$.

Check. On this line, $3x+y+3z=3t+1+3(2-t)=7$, for every $t$.

Watch out

Wrong: “The determinant is zero, so the planes form a sheaf.” Right: show that the equations are consistent and that the common set is a line. A zero determinant also occurs for a prism, parallel planes and coincident planes.

Prism arrangement

Three distinct, pairwise non-parallel planes form a prism when their pairwise intersections are three distinct parallel lines. Every pair meets, but there is no point common to all three planes. A cross-section perpendicular to the three lines gives a triangle.

Worked example 3: a contradiction gives a prism

Question. Describe the arrangement of

Plan. Compare the sum of the first two equations with the third, then identify the pairwise intersections.

These conditions contradict one another. No point can satisfy all three equations.

Why look beyond the contradiction? It establishes inconsistency, but “prism” also requires the planes to meet pairwise along different parallel lines. Here those lines are

In each line, increasing $x$ by an amount decreases $z$ by the same amount, with $y$ fixed. The directions agree, but the lines are distinct.

Answer. The planes form a prism and the system has no solution.

Check. The first line gives $2x+2z=4$ for every $t$, so none of its points lies in the third plane.

Worked example 4: two distinct parallel planes

Question. Describe the arrangement of

Plan. Divide the third equation by $2$ and compare it with the first.

Why rescale? Proportional coefficients describe parallel or coincident planes. Rescaling exposes whether the constants agree as well.

Answer. The first and third planes are distinct and parallel. The second cuts both of them. There is no common point, so the system is inconsistent.

Check. Every point of the first plane gives $2x+2y+2z=6$, never $7$, in the third equation.

Watch out

Wrong: “The planes are parallel” when the system forms a prism. Right: for a prism, the planes meet in pairs and their three intersection lines are parallel. For distinct parallel planes, the planes themselves do not meet.

Worked example 5: determinant then consistency

Question. The planes $x+y+z=3$, $x-y+2z=4$ and $2x+py+3z=q$ form a sheaf. Determine $p$ and $q$.

Plan. Set the coefficient determinant to zero, then compare the third equation with the sum of the first two.

For a sheaf, $-p=0$, hence $p=0$. Adding the first two equations now gives

so the third plane contains the common line exactly when $q=7$.

Why this second step? Setting the determinant to zero matches the coefficients, but it does not match the constants.

Answer. $p=0$, $q=7$.

Check. The third equation is exactly the sum of two non-parallel planes' equations and is distinct from both, so all three share their intersection line.

Worked example 6: a different linear combination

Question. The planes $x+y+z=3$, $x-y+2z=4$ and $2x+py+z=q$ form a sheaf. Determine $p$ and $q$.

Plan. Find the singular coefficient case, then identify which combination of the first two equations produces the third.

Thus $p=4$. Three times the first equation minus the second gives

Why this combination? Its $x$-coefficient is $3-1=2$ and its $z$-coefficient is $3-2=1$, exactly as required.

Answer. $p=4$, $q=5$.

Check. The point $(2,0,1)$ satisfies both fixed equations and gives $2(2)+4(0)+1=5$; the whole-equation combination proves the result for their entire common line.

Watch out

Wrong: in the first parameter family, $p=0$ gives a sheaf for every $q$. It only makes the coefficient determinant zero. The first two equations also require $2x+3z=7$, so the correct sheaf conditions are $p=0,\ q=7$; a different $q$ contradicts that common-line condition.

Exam tip

If told that planes form a sheaf, use $\det\mathbf{A}=0$, then explicitly impose consistency. Finding the coefficient parameter alone leaves the constants parameter undetermined.