Finding a line–plane intersection with an auxiliary plane

Content updated: October 22, 2026

To find where a line meets a plane, reduce the problem to two lines in an auxiliary plane. This yields a unique intersection only when those two lines really meet at exactly one point.

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

1. Reducing the problem to plane geometry

Let $d$ be a line and $(\alpha)$ a plane. Choose $(\beta)$ containing $d$, distinct from $(\alpha)$, and intersecting $(\alpha)$ along $\Delta$. Then

Indeed, a point of $d$ already belongs to $(\beta)$. If it also lies in $(\alpha)$, it belongs to the intersection $\Delta$. Conversely, every point of $d\cap\Delta$ lies in $d\cap(\alpha)$, since $\Delta\subset(\alpha)$. The lines $d,\Delta$ lie in $(\beta)$, so their intersection can be found by plane geometry.

2. Steps and exceptional cases

1. Choose $(\beta)$ containing $d$ so that its intersection with $(\alpha)$ is easy to find.
2. Construct $\Delta=(\alpha)\cap(\beta)$ from two common points or from one common point and a parallel direction.
3. Examine $d,\Delta$ in $(\beta)$. If they meet at $M$, it is the required intersection.
4. Check that $M$ belongs to both $d$ and $(\alpha)$; for a polyhedron problem, distinguish lines from edge segments.

If $d\parallel\Delta$ and the lines are distinct, no intersection exists: $d\parallel(\alpha)$. If $d=\Delta$, the whole line $d$ lies in $(\alpha)$; an arbitrary point cannot be called their unique intersection. Avoid choosing $(\beta)=(\alpha)$, since there would be no separate intersection line for this method.

A useful auxiliary plane is often a face containing two points defining $d$, or a diagonal plane of a pyramid. There is no need for a complicated new plane if an existing face already contains $d$.

3. Worked examples

Example 1. A diagonal plane of a pyramid

Let $S.ABCD$ have a nondegenerate parallelogram base. Let $P$ be the midpoint of $SA$. Find the intersection of the line $CP$ with the plane $(SBD)$.

Choose the auxiliary plane $(SAC)$, which contains $C,P$. Let $O=AC\cap BD$; since the base is a parallelogram, $O$ is the midpoint of $AC$. The planes $(SAC),(SBD)$ intersect along $SO$.

In triangle $SAC$, $CP$ is the median from $C$ and $SO$ the median from $S$. They meet at the centroid $G$. Hence $CP\cap(SBD)=\{G\}$, with $CG:GP=2:1$. The medians are distinct, so the intersection is unique. The point $G$ belongs to both $CP$ and $SO\subset(SBD)$, as required.

Example 2. The auxiliary plane detects parallelism

In the pyramid $S.ABC$, let $M,N$ be the midpoints of $SA,SB$. Find common points of $MN$ and $(ABC)$. Choose $(\beta)=(SAB)$, whose intersection with the base is $AB$. Since $MN$ is the midsegment of triangle $SAB$, the lines $MN,AB$ are distinct and parallel. Thus $MN\cap(ABC)=\varnothing$. There is no finite intersection point.

4. Practice

1. In the pyramid $S.ABC$, find the intersection of $SA$ with $(ABC)$.
2. In the same pyramid, use $(SAB)$ to examine the intersection of $AB$ with $(ABC)$. Why is it not unique?
3. In Example 1, can $(ABC)$ be chosen as an auxiliary plane containing $CP$?

5. Common mistakes

• Choosing an auxiliary plane that does not contain the entire line.
• Claiming an intersection before comparing $d$ with the intersection line $\Delta$.
• Proving one common point and asserting uniqueness even when the line lies in the plane.
• Rejecting a point merely because it is outside an edge segment when the problem concerns a line.

6. Related lessons

• Plane intersections from two points (English lesson)
• Plane intersections using parallelism (English lesson)
• Cross-sections of pyramids (English lesson)

Original practice resource on Scribd.