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Intersections of Two Lines in Space

Set the two line parametrizations equal, solve for their parameters, and use the direction vectors to distinguish intersection, parallelism, coincidence, and skewness.

Beginner 10 min read

What You'll Learn

  • Set two parametric line equations equal using separate parameters.
  • Solve and check all three coordinate equations consistently.
  • Find and verify a point where two lines intersect.
  • Distinguish a unique intersection from coincident lines.
  • Use direction vectors to identify parallel lines and skew lines.
  • Recognize why incompatible parameter equations mean the lines do not meet.

How Two Lines Can Relate in Space

In a plane, two distinct lines either meet or are parallel. In three-dimensional space, there is another possibility: two lines can miss each other without being parallel. Such lines are called skew. To classify a pair correctly, we need to check both whether their points can match and whether their directions are parallel.

A parametric equation describes every point on a line by varying one parameter. For example, the line through \(\mathbf{p}\) with nonzero direction vector \(\mathbf{v}\) is \(\mathbf{r}=\mathbf{p}+t\mathbf{v}\), where \(t\in\mathbb{R}\). To look for a point shared by two lines, give their parameters different names and equate their position vectors.

Definition: Two lines intersect if they have at least one common point. Distinct parallel lines have parallel direction vectors but no common point. Skew lines are nonparallel lines with no common point. If two lines have parallel directions and share a point, they are the same line; this case is called coincident.

Set the Parametrizations Equal

Write the lines as

$$ L_1:\ \mathbf{r}=\mathbf{p}+t\mathbf{v}, \qquad L_2:\ \mathbf{r}=\mathbf{q}+u\mathbf{w}, \qquad t,u\in\mathbb{R}, $$

where \(\mathbf{v}\) and \(\mathbf{w}\) are nonzero direction vectors. The parameters \(t\) and \(u\) are independent: a point reached with one value of \(t\) need not correspond to the same value of \(u\). A common point exists precisely when some pair \(t,u\) makes the position vectors equal:

$$ \mathbf{p}+t\mathbf{v}=\mathbf{q}+u\mathbf{w}. $$

This vector equation is equivalent to three scalar equations, one for each coordinate. A candidate intersection is valid only if the same pair \(t,u\) satisfies all three.

Worked Example: Find an Intersection Point

Consider the lines

$$ L_1=(1,0,2)+t(1,2,-1), \qquad L_2=(3,4,0)+u(-1,0,1). $$

Equate corresponding coordinates:

$$ 1+t=3-u,\qquad 2t=4,\qquad 2-t=u. $$

The first equation gives \(t+u=2\). The second gives \(2t=4\), so \(t=2\). Substituting into the first gives \(2+u=2\), hence \(u=0\). The third equation also holds: \(2-2=0=u\). Thus the parameters are consistent in all three coordinates.

Substitute \(t=2\) into \(L_1\):

$$ (1,0,2)+2(1,2,-1) =(1+2,\ 0+4,\ 2-2) =(3,4,0). $$

Substitute \(u=0\) into \(L_2\):

$$ (3,4,0)+0(-1,0,1)=(3,4,0). $$

Both lines contain \((3,4,0)\), so this is their intersection point. The nonparallel direction vectors also show that the lines cannot coincide.

When a Solution Exists, and When It Is Unique

The coordinate equations give a direct test: if they have a solution, the corresponding position vectors are equal, so the lines share that point. Conversely, any shared point has some parameter value on each line, and those values solve the equations. Thus solving the equations is both necessary and sufficient for finding an intersection.

Theorem: For two lines with nonzero direction vectors, solving \(\mathbf{p}+t\mathbf{v}=\mathbf{q}+u\mathbf{w}\) classifies whether they have a common point. If the direction vectors are not parallel, a common point, if it exists, is unique. If the directions are parallel, the lines either have no common point or coincide.

Here is why the uniqueness and parallel-direction claims hold. Suppose two different parameter pairs, \((t_1,u_1)\) and \((t_2,u_2)\), both give common points. Then

$$ \mathbf{p}+t_1\mathbf{v}=\mathbf{q}+u_1\mathbf{w}, \qquad \mathbf{p}+t_2\mathbf{v}=\mathbf{q}+u_2\mathbf{w}. $$

Subtract the second equation from the first:

$$ (t_1-t_2)\mathbf{v}=(u_1-u_2)\mathbf{w}. $$

If the directions are not parallel, the only way a scalar multiple of \(\mathbf{v}\) can equal a scalar multiple of \(\mathbf{w}\) is for both sides to be the zero vector. Since the direction vectors are nonzero, this forces \(t_1=t_2\) and \(u_1=u_2\). Therefore there cannot be two different shared points: any intersection is unique.

If the directions are parallel, write \(\mathbf{w}=k\mathbf{v}\) for some nonzero scalar \(k\). A common point must satisfy

$$ \mathbf{p}+t\mathbf{v}=\mathbf{q}+uk\mathbf{v}, \qquad \mathbf{q}-\mathbf{p}=(t-ku)\mathbf{v}. $$

Thus a common point exists exactly when \(\mathbf{q}-\mathbf{p}\) is a scalar multiple of \(\mathbf{v}\). If it is, the second line lies on the first and the lines coincide; otherwise, they are distinct parallel lines.

Using Direction Vectors to Complete the Classification

The cross product provides a test for whether two nonzero vectors are parallel: \(\mathbf{v}\times\mathbf{w}=\mathbf{0}\) exactly when they are parallel. This test does not, by itself, show that the lines intersect. The coordinate equations must still be checked.

Use the following conclusions after comparing the directions and solving the equations:

  • If the equations have a solution and the directions are not parallel, the lines intersect at exactly one point.
  • If the directions are parallel and the equations have no solution, the lines are distinct and parallel.
  • If the directions are parallel and the equations have a solution, the lines coincide.
  • If the directions are not parallel and the equations have no solution, the lines are skew.

Worked Example: Identify Skew Lines

Consider

$$ L_1=(0,1,2)+t(1,1,0), \qquad L_2=(2,0,1)+u(0,1,1). $$

Equating coordinates gives

$$ t=2,\qquad 1+t=u,\qquad 2=1+u. $$

The first equation requires \(t=2\). The second then requires \(u=1+2=3\). But the third requires \(2=1+u\), so \(u=1\). The requirements \(u=3\) and \(u=1\) conflict. There is no common point.

Now check that the directions are not parallel:

$$ (1,1,0)\times(0,1,1) = (1,-1,1)\ne\mathbf{0}. $$

The lines do not intersect and their directions are not parallel, so they are skew. The failed coordinate equations establish that there is no intersection; the cross product rules out parallelism.

Worked Example: Distinct Parallel Lines

Consider

$$ L_1=(1,0,2)+t(2,-1,3), \qquad L_2=(0,1,0)+u(4,-2,6). $$

The second direction vector is twice the first:

$$ (4,-2,6)=2(2,-1,3). $$

So the lines have parallel directions. To test for a common point, equate coordinates:

$$ 1+2t=4u,\qquad -t=1-2u,\qquad 2+3t=6u. $$

The first equation gives \(t=2u-\tfrac12\). The second gives \(t=2u-1\). These cannot both hold, since they would imply \(2u-\tfrac12=2u-1\), or \(-\tfrac12=-1\). Hence the lines do not meet.

The directions are parallel, but there is no common point. Therefore the lines are distinct and parallel. As a further check, the displacement from the first base point to the second is

$$ (0,1,0)-(1,0,2)=(-1,1,-2), $$

which is not a scalar multiple of \((2,-1,3)\): the first coordinate would require the scalar to be \(-\tfrac12\), while the second would require it to be \(-1\).

Worked Example: Recognize Coincident Lines

Consider

$$ L_1=(1,-2,0)+t(2,1,-1), \qquad L_2=(5,0,-2)+u(4,2,-2). $$

The second direction is twice the first. Also, the difference between the base points is

$$ (5,0,-2)-(1,-2,0)=(4,2,-2)=2(2,1,-1). $$

Thus the second base point is on \(L_1\). In fact, with \(u=0\), the second line gives \((5,0,-2)\), while \(t=2\) on the first gives

$$ (1,-2,0)+2(2,1,-1)=(5,0,-2). $$

The lines share a point and have parallel directions, so they coincide. To see that they describe the same line, rewrite the second parametrization using \(u\):

$$ (5,0,-2)+u(4,2,-2) =(1,-2,0)+(2+2u)(2,1,-1). $$

As \(u\) ranges over all real numbers, \(2+2u\) also ranges over all real numbers. Hence both parametrizations trace the same line.

A Common Mistake: Forcing the Parameters to Match

A frequent error is to assume that a shared point must occur at the same parameter value on each line. The parameters are independent; the intersection equations ask for some pair \((t,u)\), not necessarily \(t=u\).

For example, compare \(L_1=(0,0,0)+t(1,0,0)\) and \(L_2=(2,0,0)+u(0,1,0)\). Setting \(t=u\) would produce \((t,0,0)=(2,u,0)\), which requires both \(t=2\) and \(u=0\), apparently contradicting \(t=u\). That contradiction is an artifact of the incorrect extra assumption. Equating coordinates correctly gives \(t=2\), \(u=0\), and \(0=0\), so the lines intersect at \((2,0,0)\).

Takeaway: Give each line its own parameter. Solve all three coordinate equations together, then use the direction vectors to distinguish skew lines from parallel ones.

Why the Classification Matters

The coordinate method turns a spatial picture into a precise consistency test. A solution gives the parameters at which the lines share a point, and substitution verifies that point directly. If there is no solution, the direction vectors decide whether the lines are parallel or skew. Keeping these checks separate prevents a common logical gap: “the equations did not work” alone does not tell you which type of nonintersection occurred.

Check Your Understanding

For each question, distinguish the parameter test from the direction test where appropriate.

  1. What vector equation must be solved to find a common point of two parametrized lines?
  2. Why must the same pair of parameter values satisfy all three coordinate equations?
  3. If two direction vectors are nonparallel and the coordinate equations have no solution, how are the lines classified?
  4. What additional check distinguishes distinct parallel lines from coincident lines?
  5. Can two intersecting lines have parallel direction vectors without being coincident? Explain briefly.

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