From the general equations of the straight line $$r$$ we obtain: $$$\left.\begin{array}{rcl} \displaystyle \frac{x-a_1}{v_1} =\frac{y-a_2}{v_2} & \Longrightarrow & v_2(x-a_1) = v_1(y-a_2) \\ \displaystyle \frac{x-a_1}{v_1} =\frac{z-a_3}{v_3}& \Longrightarrow & v_3(x-a_1)=v_1(z-a_3)\end{array}\right\} \Longrightarrow \left. \\ \begin{array}{rcl} v_2\cdot x-v_1\cdot y+(a_2v_2-a_1v_2) &=&0 \\ v_3\cdot x-v_1\cdot z+(a_3v_1-a_1v_3) &=&0\end{array}\right\}$$$
Although generally these equations are written as: $$$\left. \begin{array}{rcl} Ax+By+Cz+D&=&0 \\ A'x+B'y+C'z+D'&=&0\end{array}\right\}$$$
and these are known as implicit equations of the straight line.
The straight line that goes through the $$A = (-1, 1, 3)$$ and that has $$\overrightarrow{v}=(3,-2,1)$$ as director vector, has this general equations: $$$\displaystyle \frac{x+1}{3}=\frac{y-1}{-2}=z-3$$$ If we separate and operate we have: $$$\left.\begin{array}{rcl} \displaystyle \frac{x+1}{3} &=& \frac{y-1}{-2} \\ \displaystyle \frac{x+1}{3} &=& z-3\end{array}\right\} \Rightarrow \left.\begin{array}{rcl} -2(x+1) &=& 3(y-1) \\ x+1 &=& 3(z-3) \end{array}\right\}\Rightarrow \left.\begin{array}{rcl} -2x-2 &=& 3y-3 \\ x+1 &=& 3z-9 \end{array}\right\} \\ \Rightarrow \left.\begin{array}{rcl}-2x-2-3y+3 &=& 0 \\ x+1 -3z+9&=& 0 \end{array}\right\}\Rightarrow \left.\begin{array}{rcl}-2x-3y+1 &=& 0 \\ x -3z+10&=& 0\end{array}\right\}$$$
which are the implicit equations of the straight line.
Find the parametric equations and a director vector of the straight line $$r$$ which implicit equations are: $$$r:\left\{\begin{array}{rcl} x+2y-z-3 &=&0 \\ 2x+5y+2z-4&=&0\end{array}\right.$$$ To convert the implicit equations into parametric equations we will solve the system using Cramer's method.
The steps are the following ones:
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We choose a minor of order two whose determinant is other than zero:$$$\left| \begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}\right|=5-4=1 \neq 0$$$
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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:$$$\left. \begin{array}{rcl} x+2y-z-3&=&0 \\ 2x+5y+2z-4&=&0 \end{array}\right\} \\ \left. \begin{array}{rcl} x+2y-k-3&=&0 \\ 2x+5y+2k-4&=&0 \end{array}\right\} \\ \left. \begin{array}{rcl} x+2y&=&k+3\\2x+5y &=& -2k+4\end{array}\right\}$$$
- Finally we apply Cramer's rule: $$$\displaystyle x=\frac{\left|\begin{matrix} k+3 & 2 \\ -2k+4 & 5 \end{matrix} \right|}{1}=5k+15+4k-8=7+9k$$$
$$$\displaystyle x=\frac{\left|\begin{matrix} 1 & k+3 \\ 2 & -2k+4 \end{matrix} \right|}{1}=-2k+4-2k-6=-2-4k$$$ Therefore the parametric equations are:$$$\left\{ \begin{array}{rcl} x&=& 7+9k\\ y&=& -2-4k \\ z&=&k\end{array}\right.$$$and a point and a vector of the straight line are:$$$ A=(7,-2,0) \qquad \overrightarrow{v}=(9,-4,1)$$$