Thread: Finding the Cartesian equation of the plane containing two lines

"Find the Cartesian equation of the plane containing the lines
L1 : x-1 = y/2 = (z+1)/-3
L2 : (x+1)/-2 = y-1, z = -1."

First of all, I tried writing out what was given into parametric scalar equations of L. However, I doubt that was the right thing to do as I was quite stuck with what to do next after that. I know that the Cartesian equation would be in the form: ax+by+cz = d (where d = ax0+by0+cz0) although, I am unsure with what to do, to get that equation. Is what I have done so far correct and if so, what do I from there? Or, is my approach towards this question wrong? I am moreover particularly confused as this question involves two lines, when I have generally only done questions involving one line/vector.

Originally Posted by cottontails

"Find the Cartesian equation of the plane containing the lines
L1 : x-1 = y/2 = (z+1)/-3
L2 : (x+1)/-2 = y-1, z = -1."

First of all, I tried writing out what was given into parametric scalar equations of L. However, I doubt that was the right thing to do as I was quite stuck with what to do next after that. I know that the Cartesian equation would be in the form: ax+by+cz = d (where d = ax0+by0+cz0) although, I am unsure with what to do, to get that equation. Is what I have done so far correct and if so, what do I from there? Or, is my approach towards this question wrong? I am moreover particularly confused as this question involves two lines, when I have generally only done questions involving one line/vector.

Finding the parametric equations of the lines would be a good start. You will then know the direction vector for each line. The cross product of those vectors will be perpendicular to the plane, so it will be proportional to the vector (a,b,c) formed be the coefficients in the equation of the plane. Finally, get the constant d by plugging in the coordinates of some point on one of the lines.