Finding plane intersections using parallel directions

Content updated: October 22, 2026

When two planes contain lines with the same direction, the direction of their intersection may be known in advance. Combining it with a common point finds the intersection without extending edges to seek a second point.

On this page: Idea · Method · Examples · Practice · Further study

1. First establish that the planes intersect

Suppose distinct planes $(\alpha),(\beta)$ intersect along $\Delta$. If they respectively contain parallel lines $d,d^{\prime}$, then $\Delta$ has their direction: it is parallel to or coincides with each, depending on position.

Do not omit the intersection hypothesis. Distinct parallel planes may contain parallel lines but have no intersection line. Coincident planes intersect in a plane, not a unique line.

2. Construction from a common point and direction

A common point $A$ is usually given. Through $A$, construct $\ell$ with the direction of $d$. Since $d\subset(\alpha)$ and $A\in(\alpha)$, the line $\ell$ lies in $(\alpha)$; if $A\in d$, then $\ell=d$. Similarly, $d^{\prime}\subset(\beta)$ and $A\in(\beta)$ place that same line $\ell$ in $(\beta)$.

The distinct planes share $\ell$, so their intersection is $\ell$. This argument establishes both a point and a direction, rather than relying on a drawing.

1. Find a common point $A$ and prove that the planes are distinct.
2. Find $d\subset(\alpha)$ and $d^{\prime}\subset(\beta)$ with the same direction.
3. Construct the line through $A$ with that direction and identify it as the intersection.
4. If $A$ lies on one of the given lines, state coincidence correctly rather than distinct parallelism.

3. Worked examples

Example 1. Opposite lateral faces of a pyramid

Let $S.ABCD$ be a pyramid with parallelogram base $ABCD$. Find the intersection of $(SAB)$ and $(SCD)$. Both contain $S$. They are distinct: a common plane would contain the distinct parallel lines $AB,CD$, hence be the base plane and contain $S$, a contradiction.

Since $AB\subset(SAB)$, $CD\subset(SCD)$, and $AB\parallel CD$, their intersection is the line through $S$ parallel to $AB$ and $CD$. It cannot coincide with a base edge because $S$ lies outside the base.

Example 2. Using a midsegment

Let $ABCD$ be a tetrahedron, with $M,N$ the respective midpoints of $AB,AC$. Find $(DMN)\cap(DBC)$. In triangle $ABC$, $MN\parallel BC$. The two planes share $D$. They are distinct because $M$ belongs to $AB$ but not to $BC=(ABC)\cap(DBC)$.

Through $D$, construct $\ell\parallel BC$. Since $MN\subset(DMN)$ and $BC\subset(DBC)$, the line $\ell$ is the intersection. Do not seek $MN\cap BC$, since those lines are distinct and parallel.

Example 3. Why is a common point needed?

Imagine horizontal planes at different heights, each containing a line with the same direction. The lines are parallel, but the planes share no point and have no intersection line. Containing parallel lines supplies a direction, but does not supply the position or existence of an intersection.

4. Practice

1. Let $S.ABCD$ be a pyramid with parallelogram base. Find $(SAD)\cap(SBC)$.
2. Two distinct planes both contain $d$. Is their intersection another line parallel to $d$?
3. In Example 2, does the intersection pass through the midpoint $M$ of $AB$?

5. Common mistakes

• Failing to check distinctness and a common point of the planes.
• Constructing a line with the correct direction but not through the common point.
• Extending parallel lines to invent a finite intersection point.
• Ignoring an intersection line that coincides with a given line.

6. Related lessons

• Plane intersections from two common points (English lesson)
• Line–plane intersections (English lesson)
• Lines parallel to planes (English lesson)
• Cross-sections (English lesson)

Original practice resource on Scribd.