Mathematics · Ch 6 — Applications of Vector Algebra

Condition for Coplanarity of Two Lines

6.8.8

Vector form. Two non-parallel lines $\vec r=\vec a+s\vec b$ and $\vec r=\vec c+t\vec d$ are coplanar exactly when they lie in a single plane. Let $A,C$ (position vectors $\vec a,\vec c$ ) be points on the two lines. Since $\vec b,\vec d$ are both parallel to that common plane, $\vec b\times\vec d$ is perpendicular to the plane; hence $\vec{AC}=\vec c-\vec a$ , which also lies in the plane, must be perpendicular to $\vec b\times\vec d$ . This gives the coplanarity condition $(\vec c-\vec a)\cdot(\vec b\times\vec d)=0.$
Cartesian form. For lines $\dfrac{x-x_1}{b_1}=\dfrac{y-y_1}{b_2}=\dfrac{z-z_1}{b_3}$ and $\dfrac{x-x_2}{d_1}=\dfrac{y-y_2}{d_2}=\dfrac{z-z_2}{d_3}$ , the same condition becomes the vanishing determinant $\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\ b_1&b_2&b_3\\ d_1&d_2&d_3\end{vmatrix}=0.$ …