Skill: Determine whether lines are parallel, intersecting, or skew

In three-dimensional space, compare two things:

1. Direction vectors tell you whether the lines point in the same direction.
2. Parametric equations tell you whether the lines share a point.

For lines

$$L_1:\mathbf r=\mathbf a+s\mathbf u \qquad\text{and}\qquad L_2:\mathbf r=\mathbf b+t\mathbf v,$$

use this classification:

• If $\mathbf u$ and $\mathbf v$ are proportional, the lines are parallel, provided they are distinct.
• If the direction vectors are not proportional, set the coordinates equal and solve for the parameters. A solution means the lines intersect.
• If the direction vectors are not proportional and the equations have no solution, the lines are skew.

Example

Determine the relationship between

$$L_1:(x,y,z)=(1,0,2)+s(1,2,1)$$

and

$$L_2:(x,y,z)=(0,1,0)+t(2,1,0).$$

First, write their coordinate equations:

For $L_1$:

$$x=1+s,\qquad y=2s,\qquad z=2+s.$$

For $L_2$:

$$x=2t,\qquad y=1+t,\qquad z=0.$$

Their direction vectors are

$$\mathbf u=(1,2,1) \qquad\text{and}\qquad \mathbf v=(2,1,0).$$

These vectors are not proportional, so the lines are not parallel. They might intersect or be skew.

To check, set corresponding coordinates equal:

$$1+s=2t,$$ $$2s=1+t,$$ $$2+s=0.$$

From the third equation,

$$s=-2.$$

Substitute this into the first equation:

$$1+(-2)=2t \quad\Rightarrow\quad t=-\frac12.$$

Now test these values in the second equation:

$$2(-2)=1+\left(-\frac12\right).$$

This gives

$$-4\ne \frac12,$$

so the equations are inconsistent. The lines have no common point.

Therefore, the lines are skew: they are not parallel, do not intersect, and do not lie in the same plane. In two-dimensional geometry, skew lines cannot occur, but they can occur in three-dimensional space.