Intersection of 3D lines (HL)

Last edited: 2026-06-09 changed to proper vector notations

There are 4 cases:

  1. Identical lines
  2. Parallel lines
  3. Intersecting lines
  4. Skew lines (ie lines on parallel planes)

Let’s say we are asked to find the intersection(s), if any, of the lines

r1=x1y1z1+λlmn\bm{r_1} = \cvec{x_1}{y_1}{z_1} + \lambda \cvec lmn
r2=x2y2z2+μpqs\bm{r_2} = \cvec{x_2}{y_2}{z_2} + \mu \cvec pqs

If l,m,n\rvec {l,m,n} is a scalar multiple of p,q,s\rvec {p,q,s}, and

  • if (x1,y1,z1)(x_1, y_1, z_1) is on line 2 (or if (x2,y2,z2)(x_2, y_2, z_2) is on line 1), then the two lines are identical
  • otherwise, the two lines are parallel

Otherwise it’s either intersecting or skew.

At the point(s) of intersection:

x:  lλ+x1=pμ+x2x:\ \ l\lambda + x_1 = p\mu + x_2
y:  mλ+y1=qμ+y2y:\ \ m\lambda + y_1 = q\mu + y_2
z:  nλ+z1=sμ+z2z:\ \ n\lambda + z_1 = s\mu + z_2

We pick two and solve for λ\lambda and μ\mu.

  • If they hold for the third equation, then there is an intersection! Substitute either λ\lambda or μ\mu into the original equations to find the point of intersection
  • otherwise, they are skew lines

There are other methods to find the point of intersection, but they involve cross products and/or magnitudes, hence larger numbers, so they are arguably harder. 

Sometimes, the two lines are linear trajectories. These typically involve time tt as a parameter.

x:  lt+x1=pt+x2x:\ \ lt + x_1 = pt + x_2
y:  mt+y1=qt+y2y:\ \ mt + y_1 = qt + y_2
z:  nt+z1=st+z2z:\ \ nt + z_1 = st + z_2

See distances between lines and between trajectories for finding the distance between the nearest approches of two linear trajectories.