Perpendicular directions of lines in space

2026-10-22 huynhvietkhanh Comments: 0

Test perpendicular directions with dot products, prove the regular-tetrahedron result, and distinguish orthogonal directions from intersection.

High schoolSolid Geometry & Vectors

Content updated: October 22, 2026

In this lesson, lines in space are called perpendicular when the angle between their directions is a right angle. They may intersect or be skew; if an intersection is also required, check it separately.

Learning goals: prove perpendicularity with direction vectors and dot products, apply the criterion to a regular tetrahedron, and check nonzero conditions.

Prerequisites: angles between lines, the dot product, and vector addition.

1. A right angle between two directions

Choose a direction vector $\vec u\ne\vec0$ for $d_1$ and $\vec v\ne\vec0$ for $d_2$. The angle between the lines belongs to $[0^\circ;90^\circ]$. It is $90^\circ$ if and only if the two direction vectors have angle $90^\circ$. Thus

$$d_1\perp d_2\quad\Longleftrightarrow\quad\vec u\cdot\vec v=0.$$

The nonzero condition is necessary: $\vec0\cdot\vec v=0$ does not determine a line direction. Reversing a direction vector or multiplying it by a number different from $0$ preserves the condition that the dot product equals $0$.

New schematic: the dot-product criterion for two nonzero direction vectors.

2. Choose vectors for the calculation

Use the difference of two distinct points on a line, or read a direction vector from parametric equations. In synthetic geometry, express vectors in terms of edges and use the given lengths and angles. Do not infer perpendicularity merely because two strokes appear to form a right angle in an image.

Using $\overrightarrow{AB}$ and $\overrightarrow{CD}$ requires $A\ne B$ and $C\ne D$. An angle question alone does not require finding an intersection. If the problem requires the lines both to intersect and to be perpendicular, check both properties.

A useful identity for any four points is

$$2\overrightarrow{AB}\cdot\overrightarrow{CD}=AD^2+BC^2-AC^2-BD^2.$$

Expand squared coordinate differences, or place the vectors at a common origin and use the dot product, to derive this identity. It calculates the product of vectors with different initial points; the four points need not be coplanar.

3. Worked examples

Example 1. Opposite edges of a regular tetrahedron

Let $ABCD$ be a regular tetrahedron with edge length $a>0$. Prove that the opposite-edge pairs $AB,CD$, $AC,BD$, and $AD,BC$ have perpendicular directions. For a geometric solution, write $\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}$. Since face $ACD$ is equilateral, $\overrightarrow{AC}$ and $\overrightarrow{CD}$ have angle $120^\circ$; since face $BCD$ is equilateral, $\overrightarrow{CB}$ and $\overrightarrow{CD}$ have angle $60^\circ$.

Therefore

$$\begin{aligned}\overrightarrow{AB}\cdot\overrightarrow{CD}&=a^2\cos120^\circ+a^2\cos60^\circ,\\&=-a^2/2+a^2/2=0.\end{aligned}$$

Hence $AB\perp CD$.

For all three opposite-edge pairs, apply the identity in Section 2 and permute the vertex names. Each right-hand side has two positive $a^2$ terms and two negative $a^2$ terms, so it equals $0$. All edges have lengths different from $0$, proving perpendicularity for each pair. The opposite-edge lines are skew, so they do not intersect.

New schematic: all six tetrahedron edges have the same length by hypothesis.

Example 2. A parameter in direction vectors

Two lines have direction vectors $(1;m;2)$ and $(2;-1;1)$, both nonzero for every $m$. Their dot product is $2-m+2=4-m$. Thus they are perpendicular exactly when $m=4$, regardless of their base points.

Example 3. Perpendicular directions without intersection

Let $d_1:(x;y;z)=(t;0;0)$ and $d_2:(x;y;z)=(0;s;1)$, with $t,s\in\mathbb R$. The direction vectors $(1;0;0)$ and $(0;1;0)$ have product $0$, so the lines are perpendicular under the stated convention. They do not intersect, since their $z$ coordinates are always $0$ and $1$, respectively.

4. Practice

Two lines have direction vectors $(2;-1;1)$ and $(1;3;1)$. Are their directions perpendicular?
Find $m$ such that lines with direction vectors $(m;1;0)$ and $(2;-1;3)$ are perpendicular.
Prove that the base diagonals $AC,BD$ of the cube $ABCD.EFGH$ with edge length $a>0$ are perpendicular.

Hints and answers

Yes, since $2-3+1=0$ and both direction vectors are nonzero.
The dot product is $2m-1$. Hence $m=1/2$; the direction vectors are always nonzero.
Set $\vec u=\overrightarrow{AB}$ and $\vec v=\overrightarrow{AD}$; they are perpendicular and both have length $a$. Then $\overrightarrow{AC}=\vec u+\vec v$ and $\overrightarrow{BD}=\vec v-\vec u$, so their dot product is $\|\vec v\|^2-\|\vec u\|^2=0$.

5. Common mistakes

Do not use the zero vector as a direction vector or replace an angle between vectors by their dot product in cosine notation. Distinguish perpendicularity of directions from the requirement that the lines have a common point.

6. Further study and sources

Vector angles and line angles in space (English lesson) · The dot product and its applications (English lesson) · Calculating the angle between lines in space (English lesson) · Skew lines (English lesson).

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