Condition for Coplanar Lines

Two lines r = (x₁,y₁,z₁)+λ(a₁,b₁,c₁) and r = (x₂,y₂,z₂)+μ(a₂,b₂,c₂) are coplanar iff this determinant vanishes. Equivalently: (a₂−a₁)·(b₁×b₂) = 0.

Two lines are coplanar iff they lie in a common plane. This happens when they either:

1. Intersect at a point, or
2. Are parallel.

Both cases correspond to the shortest distance between them being zero.

From the skew-distance formula, $d = 0$ iff:

$$(\mathbf{a_2}-\mathbf{a_1})\cdot(\mathbf{b_1}\times\mathbf{b_2}) = 0$$

Determinant form: Writing $\mathbf{a_2}-\mathbf{a_1} = (x_2-x_1, y_2-y_1, z_2-z_1)$, $\mathbf{b_1} = (a_1,b_1,c_1)$, $\mathbf{b_2} = (a_2,b_2,c_2)$:

$$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$

Geometric meaning: The three vectors $(\mathbf{a_2}-\mathbf{a_1})$, $\mathbf{b_1}$, $\mathbf{b_2}$ are coplanar (their scalar triple product is zero). This means the vector joining the two base points lies in the plane spanned by the two direction vectors — exactly the coplanarity condition.

Finding the common plane: When the lines are coplanar and intersecting, the plane contains both lines. Its normal is $\mathbf{b_1}\times\mathbf{b_2}$, and it passes through $A_1$:

$$(\mathbf{r}-\mathbf{a_1})\cdot(\mathbf{b_1}\times\mathbf{b_2}) = 0$$