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Parallel lines in 3D: distinguishing them from coincident lines

2026-10-22 huynhvietkhanh Comments: 0

Test parallel directions and point membership to distinguish distinct parallel lines, coincident lines, and skew lines.

High school3D Coordinate Geometry / Solid Geometry & Vectors

Updated: October 22, 2026

Lines with parallel direction vectors need not be distinct parallel lines: they may coincide. In Oxyz, first compare their directions, then test a point to distinguish these cases.

Learning goals: recognize distinct parallel lines; use vector tests without dividing components; distinguish parallel, coincident, and skew lines.

Prerequisites: nonzero direction vectors, the cross product, and parametric equations.

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

1. Parallel directions alone do not establish distinct parallel lines

Let $d_1$ pass through $M_1$ with direction vector $\vec u_1\ne\vec0$, and $d_2$ through $M_2$ with direction vector $\vec u_2\ne\vec0$. In this lesson, $d_1\parallel d_2$ means distinct lines with parallel directions and no common point. Coincident lines are treated separately.

If $\vec u_1,\vec u_2$ are parallel, the lines are either distinct and parallel or coincident. Test a point of one line on the other: if $M_2\in d_1$, they coincide; if $M_2\notin d_1$, they are distinct and parallel.

Distinct parallel linesTwo distinct lines in one plane have parallel direction vectors and no common point.d₁d₂u₁u₂Parallel directions; distinct lines
New schematic of two distinct lines with parallel directions.

2. Vector tests for direction and position

Set $\vec w=\overrightarrow{M_1M_2}$. The following criterion avoids coordinate fractions:

$$d_1\parallel d_2\quad\Longleftrightarrow\quad\begin{cases}[\vec u_1,\vec u_2]=\vec0,\\{}[\vec w,\vec u_1]\ne\vec0.\end{cases}$$

The first condition says that the direction vectors are parallel. The second says that $M_2$ does not lie on $d_1$: points of $d_1$ have the form $M_1+t\vec u_1$. If the second cross product is also $\vec0$, the lines coincide.

To see why testing one point suffices, suppose the lines have parallel directions and a common point $I$. Both are the set of points $I+t\vec u_1$, with $t\in\mathbb R$, so they coincide. Thus distinct lines with parallel directions cannot intersect.

State a point and direction vector for each line, check that each direction differs from $\vec0$, then examine the two cross products. This method still works when one or two direction-vector components are $0$; ratios with denominator $0$ are unnecessary.

3. Worked examples

Example 1. Checking direction and point membership

Consider the lines

$$d_1:\begin{cases}x+y+z-5=0,\\x-3y+6=0,\end{cases}\qquad d_2:\begin{cases}x=1-3t,\\y=-1-t,\\z=1+4t,\end{cases}\quad t\in\mathbb R.$$

The normal vectors of the planes defining $d_1$ are $(1;1;1)$ and $(1;-3;0)$. Their cross product is $(3;1;-4)\ne\vec0$, so take $\vec u_1=(3;1;-4)$. For $d_2$, choose $M_2(1;-1;1)$ and $\vec u_2=(-3;-1;4)=-\vec u_1$.

Substituting the parametric equations of $d_2$ into the first plane equation gives

$$(1-3t)+(-1-t)+(1+4t)-5=-4\ne0.$$

Thus no point of $d_2$ lies in the first plane, so $d_1$ and $d_2$ do not intersect. Their direction vectors are parallel, hence $d_1\parallel d_2$. There is no need to reverse the parameter sign when substituting.

Example 2. Parallel directions but coincident lines

Let $d_1$ pass through $O(0;0;0)$ with direction vector $(1;2;-1)$, and $d_2$ through $B(2;4;-2)$ with direction vector $(-2;-4;2)$. The direction vectors are parallel and $\overrightarrow{OB}=2(1;2;-1)$, so $B\in d_1$. Thus the lines coincide; they are not distinct parallel lines.

Example 3. Nonintersecting lines may be skew

The lines $d_1:(x;y;z)=(t;0;0)$ and $d_2:(x;y;z)=(0;s;1)$, with $t,s\in\mathbb R$, do not intersect because their $z$ coordinates are $0$ and $1$, respectively. However, their directions $(1;0;0)$ and $(0;1;0)$ are not parallel. The lines are skew. Nonintersection alone does not imply parallelism.

4. Practice

  1. Classify the pair $d_1:(x;y;z)=(t;0;0)$ and $d_2:(x;y;z)=(s;2;0)$.
  2. Classify the pair $d_1:(x;y;z)=(t;1+t;2)$ and $d_2:(x;y;z)=(2+3s;3+3s;2)$.
  3. Classify the pair $d_1:(x;y;z)=(t;0;0)$ and $d_2:(x;y;z)=(0;1;s)$. All parameters range over $\mathbb R$.
Hints and answers
  1. The lines are distinct and parallel: they share direction vector $(1;0;0)$, but $(0;2;0)$ does not lie on $d_1$.
  2. The lines coincide: the second direction vector is three times the first, and $(2;3;2)$ lies on $d_1$ when $t=2$.
  3. The lines are skew: their different $y$ coordinates rule out intersection, and their directions $(1;0;0)$ and $(0;0;1)$ are not parallel.

Common mistakes: concluding parallelism solely from parallel direction vectors; calling nonintersecting lines parallel without comparing directions; dividing a coordinate by $0$; using one common parameter value to test intersection of independently parametrized lines.

5. Further study and resources

Direction vectors of lines (English lesson) · Equations of lines in space (English lesson) · Intersecting lines (English lesson) · Skew lines (English lesson).

Original resource on Scribd (external link).

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