Plane Intersections: Row Reduction and Geometry

Nablio · 20 minPublished October 2, 2026

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Before you start: review linear equations and coordinates in space. This original lesson connects systems to geometry in Concordia MATH 204; it does not cover the whole course.

From two planes to a line: row reduction with geometric meaning

Find every point shared by two planes, then explain why the answer contains a free parameter. This worked example connects an augmented matrix to a line in three-dimensional space. The practice questions then compare a line of intersection, coincident planes and parallel planes.

1. Start with the geometric question

Let $x,y,z$ be the real coordinates of a point $Q=(x,y,z)$ . The two planes, named $P_1$ and $P_2$ , are $\begin{aligned}P_1:\ x+y+z&=4,\\P_2:\ 2x+y-z&=5.\end{aligned}$ We want all points that satisfy both equations at once. Each equation describes a plane because at least one of its variable coefficients is nonzero. Its coefficients form a normal vector, a nonzero vector perpendicular to that plane: $\mathbf n_1=(1,1,1)$ and $\mathbf n_2=(2,1,-1)$ . These normals are not scalar multiples, so the planes have different orientations and intersect in a line. Row reduction will locate that line.

2. Record both planes in one augmented matrix

Keep the column order $x,y,z$ throughout. Each row records one equation; the vertical divider separates the three variable coefficients from the right-hand constant. The augmented matrix is $\left[\begin{array}{rrr|r}1&1&1&4\\2&1&-1&5\end{array}\right].$ Here $R_1$ and $R_2$ mean the current first and second rows. The symbol $\leftarrow$ means “replace by.” An operation acts on the entire row, including its constant.

3. Eliminate while keeping the same common points

First use $R_2\leftarrow R_2-2R_1$ : $\left[\begin{array}{rrr|r}1&1&1&4\\0&-1&-3&-3\end{array}\right].$ The new second row says $-y-3z=-3$ . We retained the first equation, so adding twice that equation back recovers the original second equation. This reversibility preserves exactly the common solutions, even though the individual second plane changes. Next use $R_2\leftarrow -R_2$ : $\left[\begin{array}{rrr|r}1&1&1&4\\0&1&3&3\end{array}\right].$ Finally use $R_1\leftarrow R_1-R_2$ : $\left[\begin{array}{rrr|r}1&0&-2&1\\0&1&3&3\end{array}\right].$ The two remaining equations are $x-2z=1$ and $y+3z=3$ .

4. Explain why one coordinate is free

A pivot is the leading nonzero entry of a nonzero row in this reduced matrix. The pivots lie in the $x$ - and $y$ -columns; the $z$ -column has no pivot. The equations determine $x$ and $y$ once $z$ is chosen, but place no further restriction on $z$ . Choose any real number $t$ and set $z=t$ . The symbol $t$ is a parameter that labels the solutions; it does not represent time here. Then $\begin{gathered}x=1+2t,\\y=3-3t,\\z=t,\\t\in\mathbb R.\end{gathered}$ The notation $t\in\mathbb R$ means that $t$ may be any real number. Setting $t=0$ gives one point, not the complete answer.

5. Read the parameterization as a line

Combine the three coordinates into $\begin{gathered}\mathbf r(t)=(1,3,0)+t(2,-3,1),\\t\in\mathbb R.\end{gathered}$ Here $\mathbf r(t)$ is the position vector of a point on the intersection line, measured from the origin. The point $A=(1,3,0)$ is obtained at $t=0$ . The nonzero vector $\mathbf d=(2,-3,1)$ gives the line’s direction: increasing $t$ by 1 changes the coordinates by $+2,-3,+1$ . For example, $t=1$ gives $B=(3,0,1)$ . Allowing all real values of $t$ traces the whole line in both directions.

Portions of the planes x+y+z=4 and 2x+y−z=5 share the line through A=(1,3,0) and B=(3,0,1). The coordinate axes are labeled x, y and z. — Parallel projection. A and B lie in both planes, and B−A=(2,−3,1) is the line’s direction. The coloured patches show only portions of the infinite planes; their drawn edges are not boundaries of the planes. The line continues in both directions.

6. Check the whole line in the original equations

Substitute the parameterized coordinates, keeping $t$ arbitrary. For $P_1$ , $(1+2t)+(3-3t)+t=4.$ For $P_2$ , $2(1+2t)+(3-3t)-t=5.$ Both identities hold for every real $t$ , so every point on the proposed line lies in both planes. Conversely, every common point must satisfy the reduced equations, which force exactly these formulas after setting $t=z$ . This proves that the line contains all solutions. There is also a useful direction check: $\begin{aligned}\mathbf n_1\cdot\mathbf d&=2-3+1=0,\\\mathbf n_2\cdot\mathbf d&=4-3-1=0.\end{aligned}$ The dot product multiplies corresponding components and adds them. These zero results confirm that the direction is perpendicular to both normals, as a line lying in both planes requires.

7. Compare coincident planes: two free parameters

Now consider $x+y+z=4$ and $2x+2y+2z=8$ . Subtracting twice the first row from the second gives $\left[\begin{array}{rrr|r}1&1&1&4\\0&0&0&0\end{array}\right].$ The second equation contributes no new restriction: it is twice the first. These are dependent equations, and the two planes coincide. Choose $y=s$ and $z=t$ , where $s$ and $t$ are independently chosen real parameters. Then $\begin{gathered}(x,y,z)=(4-s-t,s,t),\\s,t\in\mathbb R.\end{gathered}$ Two coordinates remain free, so the common points fill a plane. A zero row does not automatically mean a line; count the free variables after checking that the system is consistent, meaning it has at least one solution.

8. Compare parallel planes: a contradiction

Change only the second constant: $x+y+z=4$ and $2x+2y+2z=9$ . The same row operation gives $\left[\begin{array}{rrr|r}1&1&1&4\\0&0&0&1\end{array}\right].$ The final row says $0=1$ , which is impossible. Doubling the first equation gives a right-hand side of 8, not 9. The normals are parallel, but the planes occupy different positions, so there is no common point. This is an inconsistent system. Compare the constants as well as the coefficients when deciding whether planes coincide.

9. Use the geometry as a final check

For exactly two genuine planes in three-dimensional space, there are three possibilities: nonparallel planes meet in a line; coincident planes share an entire plane; distinct parallel planes have no common point. They cannot have just one common point. In the matrix, these cases correspond respectively to a consistent system with one free variable, a consistent system with two free variables, and a contradiction. These conclusions concern two plane equations; adding a third equation can change the solution set.

Three questions with explained answers

1. Guided practice: find a new intersection line

For real coordinates $x,y,z$ , find all common points of $x+y+z=3$ and $x+2y+z=5$ . Write the augmented matrix in the order $x,y,z$ , begin with $R_2\leftarrow R_2-R_1$ , and express the result using $z=t$ , where $t$ is any real number. Explain why the answer is a line and verify it in both original equations.

Show the explained solution

The augmented matrix reduces as follows: $\left[\begin{array}{rrr|r}1&1&1&3\\1&2&1&5\end{array}\right].$ Applying $R_2\leftarrow R_2-R_1$ gives $\left[\begin{array}{rrr|r}1&1&1&3\\0&1&0&2\end{array}\right].$ Next, $R_1\leftarrow R_1-R_2$ gives $\left[\begin{array}{rrr|r}1&0&1&1\\0&1&0&2\end{array}\right].$ Thus $x+z=1$ and $y=2$ . There is no pivot in the $z$ -column, so choose $z=t$ . The complete solution is $\begin{gathered}(x,y,z)=(1-t,2,t)\\=(1,2,0)+t(-1,0,1),\\t\in\mathbb R.\end{gathered}$ The direction $(-1,0,1)$ is nonzero and one real parameter remains, giving a line. For every $t$ , the original left-hand sides become $(1-t)+2+t=3$ and $(1-t)+2(2)+t=5$ . Both checks hold.

2. Compare cases: when do the planes coincide?

Consider $x-2y+z=1$ and $3x-6y+3z=k$ , where $x,y,z$ are real coordinates and $k$ is a fixed real number specifying the second plane. Determine the common points for $k=3$ , then for $k\ne3$ . When there are solutions, describe all of them and state how many free parameters are needed.

Show the explained solution

Subtract three copies of the first row from the second: $\left[\begin{array}{rrr|r}1&-2&1&1\\3&-6&3&k\end{array}\right].$ The resulting matrix is $\left[\begin{array}{rrr|r}1&-2&1&1\\0&0&0&k-3\end{array}\right].$ If $k=3$ , the last row is $0=0$ and the planes coincide. Choose any real numbers $s,t$ independently, set $y=s$ , $z=t$ , and solve the first equation for $x$ : $\begin{gathered}(x,y,z)\\=(1+2s-t,s,t).\end{gathered}$ The first left-hand side is $1+2s-t-2s+t=1$ ; the second is three times this, hence 3. Two free parameters describe the entire common plane. If $k\ne3$ , the last row asserts $0=k-3$ with a nonzero right-hand side, so there are no solutions. The planes are distinct and parallel. Here $k$ selects the problem; $s,t$ label its solutions when $k=3$ .

3. Extend the reasoning: let a third plane cut the line

Return to $P_1:\ x+y+z=4$ and $P_2:\ 2x+y-z=5$ . Add $P_3:\ x+y+(1+a)z=b$ , where $a,b$ are fixed real numbers specifying a third plane. Use the known intersection line $(x,y,z)=(1+2t,3-3t,t)$ to decide when all three planes share one point, a line, or no point. Do not divide by $a$ until you have checked whether it is zero. As a numerical check, solve the case $a=2,\ b=6$ .

Show the explained solution

Substituting the line into $P_3$ gives $\begin{gathered}(1+2t)+(3-3t)\\{}+(1+a)t=4+at=b,\end{gathered}$ hence $at=b-4$ . This has a direct matrix interpretation: start from the three original rows, perform the earlier operations on rows 1 and 2, then use $R_3\leftarrow R_3-R_1-R_2$ . This last notation abbreviates two successive row operations, subtracting current row 1 and then current row 2. The result is $\left[\begin{array}{rrr|r}1&0&-2&1\\0&1&3&3\\0&0&a&b-4\end{array}\right].$ If $a\ne0$ , the third row fixes $t=z=(b-4)/a$ , so there is exactly one common point, obtained by substituting that value of $t$ into the known line. If $a=0$ and $b=4$ , the third row is $0=0$ : $P_3$ is the same plane as $P_1$ , so the entire original line remains. If $a=0$ and $b\ne4$ , the third row is a contradiction and there is no common point. For $a=2,\ b=6$ , $t=1$ , giving $(3,0,1)$ . Direct checks are $3+0+1=4$ , $2(3)+0-1=5$ , and $3+0+3(1)=6$ .

For your next problem, connect three descriptions: the plane equations, the reduced rows and the geometric solution set. State each free parameter’s allowed values and check your answer in the original equations.

Original Nablio examples for reviewing course concepts, not official exam questions.

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Plane Intersections: Row Reduction and Geometry

From two planes to a line: row reduction with geometric meaning

1. Start with the geometric question

2. Record both planes in one augmented matrix

3. Eliminate while keeping the same common points

4. Explain why one coordinate is free

5. Read the parameterization as a line

6. Check the whole line in the original equations

7. Compare coincident planes: two free parameters

8. Compare parallel planes: a contradiction

9. Use the geometry as a final check

Three questions with explained answers

1. Guided practice: find a new intersection line

2. Compare cases: when do the planes coincide?

3. Extend the reasoning: let a third plane cut the line

Useful for

Work on the step you’re missing.

Free study guides and practice

Row Operations: Why the Solutions Stay the Same

Matrices and Vectors: Practice for Concordia MATH 204