Lines and Planes in Space

Axiom 1. Through any two distinct points there passes exactly one line.

Axiom 2. Through any three points not lying on the same line there passes exactly one plane.

Axiom 3. If two points of a line lie in a plane, then the entire line lies in that plane.

Axiom 4. If two distinct planes have a point in common, then they share an entire line (their line of intersection).

Let $a$ and $b$ be two distinct lines in space.

• Intersecting lines: $a$ and $b$ lie in a common plane and have exactly one point in common.
• Parallel lines ($a \parallel b$): $a$ and $b$ lie in a common plane and have no points in common.
• Skew lines: $a$ and $b$ have no points in common and do not lie in any single plane.

Two lines $a$ and $b$ are skew if and only if no plane contains both of them simultaneously.

Proof. If $a$ and $b$ intersect or are parallel, they are coplanar by definition, so they are not skew. Conversely, if every plane containing $a$ misses at least one point of $b$, the lines share no common plane — hence they are skew. $\blacktriangleleft$

The angle between two skew lines $a$ and $b$ is defined as the angle between two intersecting lines $a'$ and $b'$ that are respectively parallel to $a$ and $b$ and pass through any common point. This angle lies in $[0°,\, 90°]$.

Two lines (intersecting or skew) are perpendicular if the angle between them is $90°$.

Let $\ell$ be a line and $\alpha$ a plane.

• Line lies in the plane ($\ell \subset \alpha$): every point of $\ell$ belongs to $\alpha$.
• Line parallel to the plane ($\ell \parallel \alpha$): $\ell$ and $\alpha$ have no common points.
• Line intersects the plane: $\ell$ meets $\alpha$ at exactly one point, called the foot of the line on the plane.

A line $\ell$ is perpendicular to a plane $\alpha$ (written $\ell \perp \alpha$) if $\ell$ is perpendicular to every line in $\alpha$ that passes through the foot of $\ell$.

If a line $\ell$ is perpendicular to two distinct lines $m$ and $n$ lying in a plane $\alpha$ and passing through the foot $O = \ell \cap \alpha$, then $\ell \perp \alpha$.

Proof sketch. Let $p$ be any line in $\alpha$ through $O$. Choose points $A \in m$, $B \in n$, $C \in p$ and a point $P$ on $\ell$ above $O$. By the condition $PA \perp OA$ and $PB \perp OB$. Using congruent triangles one shows $PC = OC$ (the distances from $P$ to points on $p$ at equal distances from $O$ are equal), which forces $\angle POC = 90°$. Hence $\ell \perp p$. Since $p$ was arbitrary, $\ell \perp \alpha$. $\blacktriangleleft$

Let $\alpha$ and $\beta$ be two distinct planes.

• Parallel planes ($\alpha \parallel \beta$): $\alpha$ and $\beta$ have no points in common.
• Intersecting planes: $\alpha$ and $\beta$ share a line $\ell = \alpha \cap \beta$ (their intersection line), guaranteed by Axiom 4.

If a line $\ell$ is perpendicular to each of two planes $\alpha$ and $\beta$, then $\alpha \parallel \beta$.

Proof. Suppose $\alpha$ and $\beta$ intersect along a line $m$. Then $m$ lies in $\alpha$, so $\ell \perp m$. Also $m$ lies in $\beta$, so $\ell \perp m$. But $\ell$ intersects $m$ at a point (since $\ell \perp \alpha$ and $\ell \perp \beta$, the foot of $\ell$ on each plane would coincide), giving a contradiction unless $\alpha \parallel \beta$. $\blacktriangleleft$

A dihedral angle is formed by two half-planes (the faces) that share a common boundary line (the edge). To measure a dihedral angle, take any point $P$ on the edge and draw, in each face, a ray perpendicular to the edge at $P$. The angle between those two rays is the measure of the dihedral angle.

In a rectangular box $ABCDA'B'C'D'$ (with $AA' \parallel BB' \parallel CC' \parallel DD'$), determine whether lines $AB$ and $A'D'$ are skew, parallel, or intersecting.

Solution. Both lines are edges of the box. $AB$ lies in the bottom face $ABCD$; $A'D'$ lies in the top face $A'B'C'D'$. These two faces are parallel and distinct.

Check for a common point: $AB$ contains points with $z = 0$ (bottom face); $A'D'$ contains points with $z = h > 0$ (top face). They share no point.

Check for a common plane: any plane containing $AB$ either is the bottom face or tilts and intersects the top face along a line. The only lines of the top face in any plane containing $AB$ would be $A'B'$ (in the plane $ABB'A'$) — but $A'D'$ is not $A'B'$ unless the box is degenerate.

More directly: $AB \parallel A'B'$ (opposite edges of face $ABB'A'$) and $A'D' \not\parallel A'B'$ (they meet at $A'$), so $A'D'$ is not parallel to $AB$. Since they don’t intersect and are not parallel, $AB$ and $A'D'$ are skew lines. $\blacktriangleleft$

In the same box with square base of side $a$ and height $h = a$, find the angle between skew lines $BD$ and $A'B'$.

Solution. Translate $A'B'$ to share a point with $BD$: since $A'B' \parallel AB$, draw the line through $B$ parallel to $A'B'$, which is $BA$ itself. So the angle between $BD$ and $A'B'$ equals $\angle DBA$.

In the square base $ABCD$ with side $a$, the diagonal $BD$ has length $a\sqrt{2}$, and $\angle DBA = 45°$.

The angle between the skew lines $BD$ and $A'B'$ is $\boxed{45°}$. $\blacktriangleleft$

In a cube $ABCDA'B'C'D'$ with edge $a$, classify each pair of lines as intersecting, parallel, or skew:

(a) $AC$ and $B'D'$; (b) $AB$ and $C'D'$; (c) $AC$ and $BD'$.

A line $\ell$ passes through vertex $A$ of triangle $ABC$ and is perpendicular to the plane of the triangle. Show that $\ell$ is perpendicular to every line in the plane of $ABC$ that passes through $A$.