Recipe

Classify a pair of lines in R3

Use when: you have two lines in $\mathbb{R}^3$ , each given as $\mathbf p_0 + t\mathbf d$ , and need to know whether they're parallel, intersecting, or skew

Goal: a definite classification — parallel (same or distinct), intersecting, or skew — plus the meeting point when one genuinely exists

Taught in Lines in R3, and skew lines

Quick recipe

Test the two directions for parallelism
If parallel, test whether the two lines are actually the same line
Set the two equations equal, coordinate by coordinate
Solve any two of the three equations for $t$ and $u$
Check the candidate pair against the third, unused equation

1
Compare $\mathbf d$ and $\mathbf e$ entry by entry: they're parallel exactly when $\mathbf d = k\mathbf e$ for one single scalar $k$ that works in every slot at once. A zero in one direction's slot has to be matched by a zero in the other's — no scalar multiple of a nonzero number is ever zero.

Example

$\mathbf d=(1,0,1)$ and $\mathbf e=(0,1,-1)$ : $\mathbf d$ has a $0$ in the middle slot, but $\mathbf e$ 's middle slot is $1$ . No scalar $k$ satisfies $0=k\cdot1$ except $k=0$ , and $k=0$ would also force $\mathbf d$ 's other entries to $0$ , which they aren't. So no single $k$ works — the directions are not parallel.
- the directions ARE parallel — go to Step 2 to test whether the lines coincide.
- the directions are NOT parallel — skip Step 2 and go straight to Step 3; only intersecting or skew remain possible.
Why: two directions are parallel exactly when one is a scalar multiple of the other; ruling this out (or in) first decides which branch of the procedure to follow next.
2
Compute $\mathbf q_0 - \mathbf p_0$ (the vector between the two named points) and check whether it, too, is a scalar multiple of $\mathbf d$ .
- $\mathbf q_0-\mathbf p_0$ is a multiple of $\mathbf d$ — same line, infinitely many shared points; stop here.
- $\mathbf q_0-\mathbf p_0$ is not a multiple of $\mathbf d$ — distinct parallel lines, no shared points, ever; stop here.
Why: two parallel lines coincide exactly when the vector joining any point of one to any point of the other lies along the shared direction; otherwise they run alongside each other forever without ever touching.
3
Write $\mathbf p_0+t\mathbf d = \mathbf q_0+u\mathbf e$ as three separate scalar equations, one per coordinate, in the two unknowns $t$ and $u$ .

Example

With $\mathbf p_0=(1,2,3)$ , $\mathbf d=(1,0,1)$ , $\mathbf q_0=(2,2,4)$ , $\mathbf e=(0,1,-1)$ : $x:\ 1+t = 2 \qquad y:\ 2 = 2+u \qquad z:\ 3+t = 4-u$

Why: a shared point has to satisfy all three coordinates simultaneously for some pair $(t,u)$ — this is just that requirement written out.
4
Pick the two equations that are easiest to solve (often the ones where a variable already stands alone) and solve them together for $t$ and $u$ .

Example

The $x$ -equation gives $t=1$ immediately; the $y$ -equation gives $u=0$ immediately. No simultaneous solving even needed here.

Why: two equations in two unknowns generically pin down one $(t,u)$ pair — but with three equations total, this pair is only a candidate until the leftover equation confirms it.
5
Substitute the $(t,u)$ found in Step 4 into the one equation not yet used. If it holds, the lines meet there; if not, they don't meet anywhere.

Example

Check $z$ : left side $3+t=3+1=4$ ; right side $4-u=4-0=4$ . They agree, so $(t,u)=(1,0)$ satisfies all three coordinates at once. Substituting $t=1$ into $\mathbf r(t)$ gives the meeting point $(2,2,4)$ — the same point $u=0$ gives back in $\mathbf s(u)$ .
- the third equation holds — genuine intersection; report the point.
- the third equation fails — the directions were non-parallel with no common point, so the lines are skew.
Why: three equations in two unknowns is generically overdetermined, so agreement on two coordinates is not yet proof of a shared point — only the third equation confirms it, or rules it out.

Common mistakes

Declaring two lines parallel from matching just one or two coordinates of the direction vectors instead of checking that a single $k$ works for every slot at once.
Skipping Step 1 and jumping straight to solving equations — when the directions ARE parallel, the "solve two, check the third" method can look inconclusive for reasons that have nothing to do with intersecting versus skew.
Treating a zero entry in a direction vector as "don't care" instead of requiring the matching entry in the other direction to be zero too.
Finding a $(t,u)$ pair that satisfies two of the three equations and stopping there, without checking the third.
Reusing the same letter for both lines' parameters, which hides that each line's parameter is independent and can (and usually does) take a different value at the meeting point.