Skew lines in 3D: two tests

2026-10-22 huynhvietkhanh Comments: 0

Detect skew lines using all coordinate equations or a scalar triple product, and construct a parallel plane through one of them.

High school3D Coordinate Geometry

Updated: October 22, 2026

Skew lines have no common point and do not lie in a common plane. Test them using a system of equations, or use a scalar triple product to detect noncoplanarity.

Learning goals: prove skewness in two ways; distinguish skew lines from parallel lines; construct a plane containing one line and parallel to the other.

Prerequisites: direction vectors, cross and dot products, systems of equations, and coplanarity.

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

1. Skew lines are noncoplanar

Let $d_1$ pass through $M_1$ with direction vector $\vec u_1\ne\vec0$, and $d_2$ through $M_2$ with direction vector $\vec u_2\ne\vec0$. They are skew when no plane contains both. Equivalently, they do not intersect and their direction vectors are not parallel.

Distinct parallel lines do not intersect but are coplanar, so they are not skew. Conversely, nonparallel directions alone do not imply skewness: the lines may intersect. Check both direction and position.

New coordinate schematic of skew lines in the distinct planes z=0 and z=1.

2. Two tests

Using a system of equations: parametrize points on $d_1$ by $t$ and points on $d_2$ by $s$, with $t,s$ independent. Equate all three coordinates. If the system has no solution and the direction vectors are not parallel, the lines are skew. Retain all three equations.

Using the scalar triple product: set $\vec w=\overrightarrow{M_1M_2}$ and

$$S=[\vec u_1,\vec u_2]\cdot\vec w.$$

Then $d_1,d_2$ are skew if and only if $S\ne0$. Indeed, a common point would imply $\vec w=t\vec u_1-s\vec u_2$, so $S=0$. Parallel direction vectors give cross product $\vec0$ and $S=0$. Thus $S\ne0$ excludes both cases.

Conversely, skew lines have independent directions. If $S=0$, then $\vec w$ lies in the vector plane spanned by $\vec u_1,\vec u_2$, so $\vec w=t\vec u_1-s\vec u_2$ for some parameters. The lines would have a common point, a contradiction. Hence $S\ne0$.

The equality $S=0$ says only that the lines are coplanar; they may intersect, be distinct and parallel, or coincide. A particular position requires further checks.

3. Examples and applications

Example 1. Two proofs of skewness

$$d_1:\begin{cases}x=t,\\y=5-2t,\\z=14-3t,\end{cases}\qquad d_2:\begin{cases}x=9-4s,\\y=1+s,\\z=-1+5s,\end{cases}\quad t,s\in\mathbb R.$$

Choose $M_1(0;5;14)$, $M_2(9;1;-1)$, $\vec u_1=(1;-2;-3)$, and $\vec u_2=(-4;1;5)$. Calculation yields

$$[\vec u_1,\vec u_2]=(-7;7;-7),\qquad\vec w=(9;-4;-15),$$

$$S=-7\cdot9+7(-4)+(-7)(-15)=14\ne0.$$

Thus the lines are skew. Reversing the cross-product order or reversing $\vec w$ may change the sign of $S$, but not the condition that it differs from $0$.

For the system test, equate the coordinates:

$$\begin{cases}t+4s=9,\\2t+s=4,\\3t+5s=15.\end{cases}$$

The first two equations give $t=1,s=2$. The last would require the false equality $3+10=15$, so the system has no solution. The nonzero cross product above also differs from $\vec0$, showing that the directions are not parallel. These two conditions again establish skewness.

Example 2. A plane containing one line and parallel to the other

For two skew lines, choose $\vec n=[\vec u_1,\vec u_2]\ne\vec0$. The plane $(P)$ through $M_1$ with normal $\vec n$ contains $d_1$, since $\vec n\cdot\vec u_1=0$. Also, $\vec n\cdot\vec u_2=0$, so the direction of $d_2$ is parallel to $(P)$.

Since $\vec n\cdot\overrightarrow{M_1M_2}=S\ne0$, the point $M_2$ does not lie in $(P)$; hence $d_2$ is disjoint from and parallel to that plane. For Example 1, simplify the normal vector to $(1;-1;1)$ and write

$$(P):x-y+z-9=0.$$

The point $M_1$ satisfies this equation, whereas every point of $d_2$ makes its left-hand side $-2$. This directly verifies that $d_2$ does not meet $(P)$.

4. Practice

Classify each pair, with real parameters:

$d_1:(x;y;z)=(t;0;0)$, $d_2:(x;y;z)=(0;s;1)$.
$d_1:(x;y;z)=(t;0;0)$, $d_2:(x;y;z)=(1;0;s)$.
$d_1:(x;y;z)=(t;0;0)$, $d_2:(x;y;z)=(s;2;3)$.

Hints and answers

The scalar triple product is $1$, so the lines are skew.
The lines intersect at $(1;0;0)$. Their scalar triple product is $0$, and their direction vectors are not parallel.
The lines are distinct and parallel: their directions agree, but $(0;2;3)$ does not lie on the first line. Their scalar triple product is $0$.

Common mistakes: concluding skewness from nonintersection alone; substituting the same parameter value into both lines; discarding the third coordinate equation; inferring parallelism from $S=0$; forgetting that direction vectors must be nonzero.