Relation Between 2 Lines

Given vector equation of 2 Lines

\(\vec{L_1} = \vec{A} + a\vec{B}\)
\(\vec{L_2}= \vec{C} + c\vec{C}\)
In above equations \(\vec{L_1}\) is a Line having a point with position vector \(\vec{A}\) and direction vector \(\vec{B}\) and \(\vec{L_2}\) is a Line having a point with position vector \(\vec{C}\) and direction vector \(\vec{D}\). The following can be the relation between these 2 Lines.

  1. If \(\vec{B} \times \vec{D} = 0\) then the lines do not intersect. Further, if \((\vec{C} - \vec{A}) \times \vec{D} = 0\) (or \((\vec{C} - \vec{A}) \times \vec{B} = 0\)) then the lines are coincident. Otherwise the lines are parallel.
  2. If \(\vec{B} \times \vec{D} \neq 0\) then the lines intersect. The point of intersection is given by the following formula

    \(\vec{P} = \vec{A} + (\frac{\vert (\vec{C}-\vec{A})\times \vec{D} \vert}{\vert \vec{B}\times \vec{D} \vert}) s\vec{B}\)
    OR
    \(\vec{P} = \vec{C} + (\frac{\vert (\vec{A}-\vec{C})\times \vec{B} \vert}{\vert \vec{D}\times \vec{B} \vert}) s\vec{D}\)
    The value of s in above equations is 1 if \( ((\vec{C}-\vec{A})\times \vec{D}) \cdot (\vec{B}\times \vec{D}) > 0 \) (and \( ((\vec{A}-\vec{C})\times \vec{B}) \cdot (\vec{D}\times \vec{B}) > 0 \) ). If these dot products are < 0 then the value of s is -1.
The above formulae can be used for finding point of intersection of 2 Lines in both 2D and 3D.

In 2D only, scalar cartesian equation of two lines are given as follows

\(A_1x + B_1y = C_1\)
\(A_2x + B_2y = C_2\)

These equations can be represented in matrix form as following

\(\begin{bmatrix} A_1 & B_1 \\A_2 & B_2 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} C_1 \\C_2 \end{bmatrix}\)

The following can be the relation between these 2 Lines
  1. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\) then the lines are coincident.
  2. If \(\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\) then the lines are parallel.
  3. If \(\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\) then the lines intersect.
Following are the steps to find the point of intersection between these 2 Lines:
  1. Calulate the following determinants

    \(D=\begin{vmatrix} A_1 & B_1 \\A_2 & B_2 \end{vmatrix} \hspace{.5cm} D_1=\begin{vmatrix} C_1 & B_1 \\C_2 & B_2 \end{vmatrix} \hspace{.5cm} D_2=\begin{vmatrix} A_1 & C_1 \\A_2 & C_2 \end{vmatrix} \)
  2. If \(D=D_1=D_2=0\) then the lines are coincident.
  3. If \(D=0\) and either \(D_1\neq0\) or \(D_2\neq0\) then the lines are parallel.
  4. If \(D\neq0\) then the lines intersect and the coordinates of intersection is given as

    \(x=\frac{D_1}{D} \hspace{.5cm} y=\frac{D_2}{D}\)
  5. The point of intersection between these 2 Lines can be also be found out by the matrix inversion as following

    \(\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} A_1 & B_1 \\A_2 & B_2 \end{bmatrix}^{-1} \begin{bmatrix} C_1 \\C_2 \end{bmatrix}\)

    \(\Rightarrow \begin{bmatrix} x \\ y\end{bmatrix} = \frac{1}{D}\begin{bmatrix} B_2 & -B_1 \\-A_2 & A_1 \end{bmatrix} \begin{bmatrix} C_1 \\C_2 \end{bmatrix}\)