Implicit equations of a straight line in the space

From the general equations of the straight line $r$ we obtain: $\begin{array}{rcl} \frac{x - a_{1}}{v_{1}} = \frac{y - a_{2}}{v_{2}} & ⟹ & v_{2} (x - a_{1}) = v_{1} (y - a_{2}) \\ \frac{x - a_{1}}{v_{1}} = \frac{z - a_{3}}{v_{3}} & ⟹ & v_{3} (x - a_{1}) = v_{1} (z - a_{3}) \end{array}} ⟹ \begin{array}{rcl} v_{2} \cdot x - v_{1} \cdot y + (a_{2} v_{2} - a_{1} v_{2}) & = & 0 \\ v_{3} \cdot x - v_{1} \cdot z + (a_{3} v_{1} - a_{1} v_{3}) & = & 0 \end{array}}$

Although generally these equations are written as: $\begin{array}{rcl} A x + B y + C z + D & = & 0 \\ A^{'} x + B^{'} y + C^{'} z + D^{'} & = & 0 \end{array}}$

and these are known as implicit equations of the straight line.

Example

The straight line that goes through the $A = (- 1, 1, 3)$ and that has $\vec{v} = (3, - 2, 1)$ as director vector, has this general equations: $\frac{x + 1}{3} = \frac{y - 1}{- 2} = z - 3$ If we separate and operate we have: $\begin{array}{rcl} \frac{x + 1}{3} & = & \frac{y - 1}{- 2} \\ \frac{x + 1}{3} & = & z - 3 \end{array}} \Rightarrow \begin{array}{rcl} - 2 (x + 1) & = & 3 (y - 1) \\ x + 1 & = & 3 (z - 3) \end{array}} \Rightarrow \begin{array}{rcl} - 2 x - 2 & = & 3 y - 3 \\ x + 1 & = & 3 z - 9 \end{array}} \Rightarrow \begin{array}{rcl} - 2 x - 2 - 3 y + 3 & = & 0 \\ x + 1 - 3 z + 9 & = & 0 \end{array}} \Rightarrow \begin{array}{rcl} - 2 x - 3 y + 1 & = & 0 \\ x - 3 z + 10 & = & 0 \end{array}}$

which are the implicit equations of the straight line.

Example

Find the parametric equations and a director vector of the straight line $r$ which implicit equations are: $r : {\begin{array}{rcl} x + 2 y - z - 3 & = & 0 \\ 2 x + 5 y + 2 z - 4 & = & 0 \end{array}$ To convert the implicit equations into parametric equations we will solve the system using Cramer's method.

The steps are the following ones:

We choose a minor of order two whose determinant is other than zero: $| \begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix} | = 5 - 4 = 1 \neq 0$
We replace the unknown that does not intervene in this minor (in this case $z$ ), for a parameter $k$ . We have, therefore, $z = k$ , and we isolate the variables: $\begin{array}{rcl} x + 2 y - z - 3 & = & 0 \\ 2 x + 5 y + 2 z - 4 & = & 0 \end{array}} \begin{array}{rcl} x + 2 y - k - 3 & = & 0 \\ 2 x + 5 y + 2 k - 4 & = & 0 \end{array}} \begin{array}{rcl} x + 2 y & = & k + 3 \\ 2 x + 5 y & = & - 2 k + 4 \end{array}}$
Finally we apply Cramer's rule: $x = \frac{| \begin{matrix} k + 3 & 2 \\ - 2 k + 4 & 5 \end{matrix} |}{1} = 5 k + 15 + 4 k - 8 = 7 + 9 k$

$x = \frac{| \begin{matrix} 1 & k + 3 \\ 2 & - 2 k + 4 \end{matrix} |}{1} = - 2 k + 4 - 2 k - 6 = - 2 - 4 k$ Therefore the parametric equations are: ${\begin{array}{rcl} x & = & 7 + 9 k \\ y & = & - 2 - 4 k \\ z & = & k \end{array}$ and a point and a vector of the straight line are: $A = (7, - 2, 0) \vec{v} = (9, - 4, 1)$