Coplanarity of Two Lines

Two lines in space may or may not lie in a common plane. Parallel lines always do — any two parallel lines determine a plane between them. The interesting case is a pair of non-parallel lines: such lines are coplanar precisely when the shortest distance between them is zero (if it were positive, the lines would be skew, passing each other in different planes without ever meeting). This gives a clean test: two non-parallel lines $\vec r=\vec a_1+\lambda_1\vec b_1$ and $\vec r=\vec a_2+\lambda_2\vec b_2$ are coplanar if and only if

When this holds, the common plane can be pinned down immediately: it passes through the point $A(\vec a_1)$, and since both $\vec b_1$ and $\vec b_2$ lie in the plane, their cross product $\vec b_1\times\vec b_2$ must be normal to it. So the plane's equation is

In Cartesian form, for lines $\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}$ and $\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_2}$, the same idea becomes a $3\times3$ determinant test:

and the plane determined by the two lines is

Worked example. Show that the lines $\vec r=(\hat i+\hat j-\hat k)+\lambda(2\hat i-2\hat j+\hat k)$ and $\vec r=(4\hat i-3\hat j+2\hat k)+\mu(\hat i-2\hat j+2\hat k)$ are coplanar, and find the plane they determine.

Here $\vec a_1=\hat i+\hat j-\hat k$, $\vec a_2=4\hat i-3\hat j+2\hat k$, $\vec b_1=2\hat i-2\hat j+\hat k$, $\vec b_2=\hat i-2\hat j+2\hat k$, so $\vec a_2-\vec a_1=3\hat i-4\hat j+3\hat k$.

$(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=\begin{vmatrix}3&-4&3\\2&-2&1\\1&-2&2\end{vmatrix}=3(-4+2)-(-4)(4-1)+3(-4+2)=-6+12-6=0,$ …