Intersections of Planes

Every system of linear equations in $x$, $y$ and $z$ is secretly a question about planes: where do they all meet? Two planes usually meet in a line. Three planes can meet in a single point, along a shared line, or not have any point in common at all, in several different ways. This page shows you how to solve the system by elimination and how to read the geometry from the algebra.

Key ideas

Two planes

Two planes with normals $\vec{n}_1$ and $\vec{n}_2$:

Normals	Equations	Intersection
not parallel		a line, with direction $\vec{n}_1 \times \vec{n}_2$
parallel	not multiples of each other	none (parallel and distinct)
parallel	multiples of each other	the whole plane (coincident)

To find the line, eliminate one variable, let another be the parameter $t$, and solve for the rest, as in linear equations in 2-space and 3-space.

Three planes: all the configurations

A system of three equations in three unknowns is consistent if it has at least one solution and inconsistent if it has none.

Consistent systems (the planes share at least one point):

• One point. The usual case: like the floor and two walls meeting at a corner.
• A line, sheaf. Three different planes all pass through one common line, like pages of a book around the spine.
• A line, two coincident. Two of the planes are the same, and the third crosses it in a line.
• A plane. All three planes are the same.

Inconsistent systems (no point is on all three):

• Three parallel planes, all distinct.
• Two parallel planes and a third crossing them: the third plane cuts the other two in two parallel lines.
• Two coincident planes and a third parallel to them.
• Triangular prism. No two planes are parallel, and each pair meets in a line, but the three lines are parallel, like the three sides of a Toblerone box.

Seen edge-on (looking along the direction they share), each plane looks like a line.

Solving by elimination

1. Use pairs of equations to eliminate the same variable twice. You get two equations in two unknowns.
2. Eliminate again to get one equation in one unknown.
3. Back-substitute to find the others.

Then read the result:

What elimination gives	Meaning
a single value for each variable	one point of intersection
an equation that becomes $0 = 0$	infinitely many solutions: let a variable be $t$ (a line, or a whole plane if all three equations are multiples of one another)
a false statement like $0 = 4$	no solution: inconsistent

To tell the inconsistent cases apart, look at the normals. Two parallel normals mean two parallel planes; no parallel normals at all, in an inconsistent system, means a triangular prism. Likewise, a line of solutions with no two planes coincident is a sheaf.

Predicting with the normals

Three planes meet in exactly one point when their normals are not coplanar (they don’t all lie in one plane). You can test that with the triple product (the scalar triple product from cross product applications):

$$\vec{n}_1 \cdot (\vec{n}_2 \times \vec{n}_3)$$

• If $\vec{n}_1 \cdot (\vec{n}_2 \times \vec{n}_3) \ne 0$: exactly one point of intersection.
• If $\vec{n}_1 \cdot (\vec{n}_2 \times \vec{n}_3) = 0$: the normals are coplanar, so it’s a line (sheaf), a prism, or one of the cases with parallel planes. Use elimination to decide.

Why it works: $\vec{n}_2 \times \vec{n}_3$ is perpendicular to both $\vec{n}_2$ and $\vec{n}_3$ (see cross product). The dot product with $\vec{n}_1$ is zero exactly when $\vec{n}_1$ lies in the same plane as the other two normals.

Worked examples

Example 1: Two planes

Find the intersection of each pair of planes.

• (a) $x - 2y + 3z = 4$ and $-2x + 4y - 6z = 1$
• (b) $2x + y - z = 3$ and $x - y + 2z = 0$

Solution.

(a) $[-2, 4, -6] = -2[1, -2, 3]$, so the normals are parallel. Multiplying the first equation by $-2$ gives $-2x + 4y - 6z = -8$, not $1$. The planes are parallel and distinct: no intersection.

(b) The normals aren’t parallel, so the planes meet in a line. Add the equations to eliminate $y$:

$$3x + z = 3 \quad\Rightarrow\quad z = 3 - 3x$$

Let $x = t$. Then $z = 3 - 3t$, and from the second equation $y = x + 2z = t + 6 - 6t = 6 - 5t$.

$$\vec{r} = [0, 6, 3] + t[1, -5, -3], \quad t \in \mathbb{R}$$

Check the direction: $[2, 1, -1] \times [1, -1, 2] = [1, -5, -3]$ ✓. Check the point $(0, 6, 3)$: $0 + 6 - 3 = 3$ ✓ and $0 - 6 + 6 = 0$ ✓.

Example 2: Three planes meeting in a point

Solve the system and interpret it geometrically.

$$\begin{aligned} x + y + z &= 6 && (1) \\ 2x - y + z &= 3 && (2) \\ x + 2y - z &= 2 && (3) \end{aligned}$$

Solution. Eliminate $z$ twice:

$$\begin{aligned} (1) + (3):&\quad 2x + 3y = 8 && (4) \\ (2) + (3):&\quad 3x + y = 5 && (5) \end{aligned}$$

Now $3 \times (5) - (4)$ eliminates $y$: $9x + 3y - 2x - 3y = 15 - 8$, so $7x = 7$ and $x = 1$.

From (5): $y = 5 - 3 = 2$. From (1): $z = 6 - 1 - 2 = 3$.

The three planes meet at the single point $(1, 2, 3)$.

Check in (2) and (3): $2 - 2 + 3 = 3$ ✓ and $1 + 4 - 3 = 2$ ✓.

The triple product agrees: $[1, 1, 1] \cdot \big([2, -1, 1] \times [1, 2, -1]\big) = [1, 1, 1] \cdot [-1, 3, 5] = 7 \ne 0$.

Example 3: A sheaf of planes

Solve the system and interpret it geometrically.

$$\begin{aligned} x + y - z &= 2 && (1) \\ 2x - y + z &= 1 && (2) \\ 4x + y - z &= 5 && (3) \end{aligned}$$

Solution. Eliminate $y$ and $z$ together:

$$\begin{aligned} (1) + (2):&\quad 3x = 3 && (4) \\ (2) + (3):&\quad 6x = 6 && (5) \end{aligned}$$

Both give $x = 1$. Equation (5) is just $2 \times (4)$, so eliminating between them gives $0 = 0$: infinitely many solutions.

With $x = 1$, equation (1) gives $y - z = 1$. Let $z = t$, so $y = 1 + t$:

$$\vec{r} = [1, 1, 0] + t[0, 1, 1], \quad t \in \mathbb{R}$$

No two normals are parallel, so the planes are all different. They form a sheaf through this line. (Notice that equation (3) is $2 \times (1) + (2)$.)

Check $(1, 2, 1)$, at $t = 1$: $1 + 2 - 1 = 2$ ✓, $2 - 2 + 1 = 1$ ✓, $4 + 2 - 1 = 5$ ✓.

Example 4: A triangular prism

Change one number in Example 3: solve

$$\begin{aligned} x + y - z &= 2 && (1) \\ 2x - y + z &= 1 && (2) \\ 4x + y - z &= 3 && (3) \end{aligned}$$

Solution.

$$\begin{aligned} (1) + (2):&\quad 3x = 3 \quad\Rightarrow\quad x = 1 \\ (2) + (3):&\quad 6x = 4 \quad\Rightarrow\quad x = \tfrac{2}{3} \end{aligned}$$

$x$ can’t be both $1$ and $\frac{2}{3}$. (Equivalently, $2 \times (1) + (2)$ gives $4x + y - z = 5$, and with (3) that means $5 = 3$, which is false.) The system is inconsistent.

The normals $[1, 1, -1]$, $[2, -1, 1]$ and $[4, 1, -1]$ are not parallel in pairs, so no two planes are parallel. Each pair meets in a line, but the three lines never meet: the planes form a triangular prism.

Common mistakes

Assuming no solution means parallel planes. A triangular prism has no solution, yet no two of its planes are parallel. Check the normals before you describe the picture.

Eliminating different variables from different pairs. If you eliminate $z$ from one pair and $y$ from another, you’re left with a mix that doesn’t reduce. Eliminate the same variable from two different pairs.

Reading 0 = 0 as “no solution”. $0 = 0$ means one equation was a combination of the others. You’ve lost an equation, not the solutions: there are infinitely many, usually a line.

Trusting a zero triple product to mean “a line”. A zero triple product only says the normals are coplanar. The planes could form a sheaf (a line) or a prism or include parallel planes (no solution). Elimination decides.

Forgetting to check in all three equations. A point that satisfies two of the equations is only on two of the planes. Always substitute into the third.

Practice

1. (Warm-up) Describe the intersection of each pair of planes.

• (a) $3x - y + 2z = 1$ and $6x - 2y + 4z = 2$
• (b) $x + y + z = 1$ and $x + y + z = 4$
• (c) $x - y = 2$ and $y + z = 1$

Solution

(a) The second equation is $2$ times the first: the planes are coincident.

(b) Same normal, different constants: parallel and distinct, no intersection.

(c) The normals $[1, -1, 0]$ and $[0, 1, 1]$ aren’t parallel: the planes meet in a line.

2. (Warm-up) Verify that $(2, -1, 1)$ is on all three planes $x + y + z = 2$, $2x - y + 3z = 8$ and $x + 3y - z = -2$.

Solution

$2 - 1 + 1 = 2$ ✓, $4 + 1 + 3 = 8$ ✓, $2 - 3 - 1 = -2$ ✓. The point is on all three planes.

3. (Core) Solve and interpret: $x + 2y - z = 3$, $2x - y + z = 4$, $x + y + 2z = 5$.

Solution

Label the equations (1), (2), (3).

$$\begin{aligned} (1) + (2):&\quad 3x + y = 7 && (4) \\ 2 \times (1) + (3):&\quad 3x + 5y = 11 && (5) \end{aligned}$$

$(5) - (4)$: $4y = 4$, so $y = 1$. Then $3x = 6$ and $x = 2$. From (1): $z = 2 + 2 - 3 = 1$.

The planes meet at the single point $(2, 1, 1)$.

Check (2) and (3): $4 - 1 + 1 = 4$ ✓ and $2 + 1 + 2 = 5$ ✓.

4. (Core) Find a vector equation of the line of intersection of the planes $x + 2y + z = 5$ and $x - y + z = 2$.

Solution

Subtract the equations: $3y = 3$, so $y = 1$. Then $x + z = 3$. Let $z = t$, so $x = 3 - t$:

$$\vec{r} = [3, 1, 0] + t[-1, 0, 1]$$

Check the direction: $[1, 2, 1] \times [1, -1, 1] = [3, 0, -3] = -3[-1, 0, 1]$ ✓.

5. (Core) Consider the planes $2x + y - z = 1$, $x - y + 2z = 3$ and $3x + z = 4$.

• (a) Use the triple product of the normals to decide whether the planes meet in exactly one point.
• (b) Solve the system and describe the intersection.

Solution

(a)

$$[1, -1, 2] \times [3, 0, 1] = [(-1)(1) - 2(0),\ 2(3) - 1(1),\ 1(0) - (-1)(3)] = [-1, 5, 3]$$$$[2, 1, -1] \cdot [-1, 5, 3] = -2 + 5 - 3 = 0$$

The normals are coplanar, so the planes do not meet in exactly one point.

(b) Adding the first two equations gives $3x + z = 4$, which is exactly the third equation. So the third equation adds nothing new: infinitely many solutions. Let $x = t$. Then $z = 4 - 3t$, and from the first equation $y = 1 - 2x + z = 1 - 2t + 4 - 3t = 5 - 5t$:

$$\vec{r} = [0, 5, 4] + t[1, -5, -3]$$

No two normals are parallel, so this is a sheaf of three planes through this line.

6. (Core) Solve and interpret: $x - y + 2z = 3$, $2x - 2y + 4z = 1$, $x + y + z = 2$.

Solution

$2 \times$ the first equation is $2x - 2y + 4z = 6$, but the second says $2x - 2y + 4z = 1$. That would mean $6 = 1$, so the system is inconsistent.

The first two planes are parallel and distinct (normals $[1, -1, 2]$ and $[2, -2, 4]$). The third plane’s normal $[1, 1, 1]$ isn’t parallel to them, so it cuts both, in two parallel lines. There is no point on all three planes.

7. (Core) A bakery in Halifax sells muffins, cookies, and scones. One order of 2 muffins, 3 cookies and 1 scone costs $12.50. A second order of 1 muffin, 4 cookies and 2 scones costs $14.50. A third order of 3 muffins, 1 cookie and 2 scones costs $15.00. Find the price of each item.

Solution

Let $m$, $c$ and $s$ be the prices in dollars:

$$\begin{aligned} 2m + 3c + s &= 12.50 && (1) \\ m + 4c + 2s &= 14.50 && (2) \\ 3m + c + 2s &= 15.00 && (3) \end{aligned}$$

Eliminate $s$:

$$\begin{aligned} 2 \times (1) - (2):&\quad 3m + 2c = 10.50 && (4) \\ 2 \times (1) - (3):&\quad m + 5c = 10.00 && (5) \end{aligned}$$

$3 \times (5) - (4)$: $13c = 19.50$, so $c = 1.50$. Then $m = 10 - 7.5 = 2.50$, and from (1), $s = 12.50 - 5 - 4.50 = 3.00$.

A muffin costs $2.50, a cookie $1.50, and a scone $3.00.

Check (2): $2.50 + 6 + 6 = 14.50$ ✓. Check (3): $7.50 + 1.50 + 6 = 15.00$ ✓. Geometrically, the three planes meet in the single point $(2.5, 1.5, 3)$.

8. (Challenge) Let $P_1: x + 2y - z = 4$ and $P_2: 2x - y + z = 1$.

• (a) Show that the plane $P_3: x = 0$ meets $P_1$ and $P_2$ in exactly one point, and find it.
• (b) Write the equation of a plane $P_4$, different from $P_1$ and $P_2$, that meets them in a whole line.

Solution

(a) Triple product of the normals:

$$[2, -1, 1] \times [1, 0, 0] = [(-1)(0) - 1(0),\ 1(1) - 2(0),\ 2(0) - (-1)(1)] = [0, 1, 1]$$$$[1, 2, -1] \cdot [0, 1, 1] = 0 + 2 - 1 = 1 \ne 0$$

So the three planes meet in exactly one point. With $x = 0$: $2y - z = 4$ and $-y + z = 1$. Adding gives $y = 5$, then $z = 6$. The point is $(0, 5, 6)$.

(b) Any combination of the two equations contains their line of intersection. Adding them gives

$$P_4: 3x + y = 5$$

Every point on both $P_1$ and $P_2$ satisfies $P_4$, so the three planes form a sheaf. (Many other answers work, such as $P_1 - P_2$: $-x + 3y - 2z = 3$.)

9. (Challenge) For which values of $k$ does the system $x + y + z = 2$, $x - y + 2z = 1$, $2x + kz = 3$ have exactly one solution? Describe the intersection for the remaining value of $k$.

Solution

Triple product of the normals:

$$[1, -1, 2] \times [2, 0, k] = [(-1)k - 2(0),\ 2(2) - 1(k),\ 1(0) - (-1)(2)] = [-k,\ 4 - k,\ 2]$$$$[1, 1, 1] \cdot [-k,\ 4 - k,\ 2] = -k + 4 - k + 2 = 6 - 2k$$

This is non-zero unless $k = 3$, so there is exactly one solution for every $k \ne 3$.

For $k = 3$: adding the first two equations gives $2x + 3z = 3$, which is exactly the third equation. So there are infinitely many solutions. Let $z = 1 + 2t$; then $2x = 3 - 3z = -6t$, so $x = -3t$, and $y = 2 - x - z = 1 + t$:

$$\vec{r} = [0, 1, 1] + t[-3, 1, 2]$$

No two normals are parallel, so the three planes form a sheaf through this line.