A line perpendicular to a plane: criteria and examples

Content updated: October 22, 2026

A line perpendicular to a plane can be recognized by two independent directions within the plane. Perpendicularity to two parallel lines checks only one direction and is insufficient.

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1. Definition and perpendicular foot

A line $d$ is perpendicular to a plane $(P)$ if $d$ is perpendicular to every line in $(P)$, written $d\perp(P)$. Angles between lines in space are measured between their directions, so perpendicular lines need not intersect under this convention.

If $d\perp(P)$, it meets $(P)$ at a unique point $H$, called the foot of the perpendicular. The direction of $d$ is a normal direction to $(P)$. Thus neither $d\subset(P)$ nor $d\parallel(P)$ is possible.

2. Two criteria

The intersecting-lines criterion. If $a,b\subset(P)$ are distinct intersecting lines, $d\perp a$, and $d\perp b$, then $d\perp(P)$. Conversely, the definition implies that $d\perp(P)$ entails both perpendicularity relations.

To see why intersection suffices, choose direction vectors $\vec u,\vec v$ for $a,b$. They are independent and span all directions in $(P)$. A direction vector of $d$ is perpendicular to both $\vec u,\vec v$, so its dot product with every linear combination of them is $0$. Thus it is perpendicular to every direction in the plane.

Transfer through parallelism. If $a\parallel b$ and $a\perp(P)$, then $b\perp(P)$, since the two lines have the same normal direction. When applying the first criterion, state where the two lines in the plane intersect.

3. Worked examples

Example 1. An edge of a cube

Let $ABCD.EFGH$ be a cube, with $E,F,G,H$ above $A,B,C,D$, respectively. We have $EA\perp AB$ and $EA\perp AD$, while $AB,AD$ meet at $A$ and both lie in $(ABCD)$. Therefore $EA\perp(ABCD)$. The relations $EA\perp AB$ and $EA\perp CD$ alone do not satisfy the criterion, since $AB\parallel CD$.

Example 2. Why not use two parallel lines?

In the same cube, $BDHF$ is a rectangle, so $BD\perp DH$. By Example 1, $AE\perp(ABCD)$, implying $BD\perp AE$. The lines $AE,DH$ both lie in $(AEDH)$ but are parallel. The line $BD$ is not perpendicular to $(AEDH)$: otherwise it would be perpendicular to $AD$, whereas the diagonal $BD$ is not perpendicular to side $AD$ of the square. This is a counterexample to omitting the intersection condition.

Example 3. Transfer perpendicularity through a parallelogram

Let $S.ABC$ be a pyramid whose base triangle $ABC$ is right-angled at $A$, with $SA\perp(ABC)$. Let $M$ be the midpoint of $AC$, and $D$ the reflection of $B$ in the point $M$. Prove $CD\perp(SAC)$.

Since $AB\perp AC$ and $AB\perp SA$, and $AC,SA$ meet in $(SAC)$, we obtain $AB\perp(SAC)$. The diagonals $AC,BD$ have the same midpoint $M$, so $ABCD$ is a parallelogram. Hence $CD\parallel AB$, implying $CD\perp(SAC)$. The nondegenerate base triangle ensures that all lines and planes used here are defined.

4. Practice

1. In the cube of Example 1, prove $BF\perp(ABCD)$.
2. Suppose $d\perp a$ and $d\perp b$, where $a,b$ are distinct parallel lines in $(P)$. Can we conclude $d\perp(P)$?
3. For the pyramid in Example 3, prove $AB\perp SC$.

5. Common mistakes

• Replacing “every line in the plane” by “one line.”
• Using two parallel lines instead of two independent directions.
• Failing to check that both chosen lines lie in the plane under consideration.
• Requiring every pair of perpendicular lines in space to intersect.

6. Related lessons

• Perpendicular lines (English lesson)
• The three perpendiculars theorem (English lesson)
• Perpendicular planes (English lesson)

Original practice resource on Scribd.