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Given two lines: ℓ1 : (2, −1, −1) + t(3, 2, 0), ℓ2 : (−1, 2, −4) + s(9, 1, 3) how can i find the cartesian equation of the plane that is parallel to the two lines and passes through the point A(−2, 1, −3) without using cross product?

I have only found ways to solve this using cross product and I was wondering if it can be solved without using cross product? I used the directional vectors of the lines as directional vectors of the plane to find the first two equations and then, I used A to find another equation.

  1. 3a + 2b = 0 => the plane normal is perpendicular to ℓ1 directional vector
  2. 9a + b + 3c = 0 => the plane normal is perpendicular to ℓ2 directional vector
  3. -2a + b - 3c + d = 0 => point on the plane

so that's gives me three equations with 4 variables (a,b,c,d) of the plane parameters. how do I proceed from here? use the points from on of the lines? is there another way to do it without cross product? quicker way?

Thank you,

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  • $\begingroup$ Please use formulas to show exactly what you did. $\endgroup$
    – Kurt G.
    Commented Apr 24, 2022 at 16:47
  • $\begingroup$ I've edited my question $\endgroup$
    – Danny
    Commented Apr 24, 2022 at 17:01
  • $\begingroup$ You are already doing it without cross product. The condition that the plane normal can have an arbitrary non zero length (preferrably one) should give you the fourth equation. $\endgroup$
    – Kurt G.
    Commented Apr 24, 2022 at 17:12
  • $\begingroup$ can you please write the 4'th equation? $\endgroup$
    – Danny
    Commented Apr 24, 2022 at 17:39
  • $\begingroup$ You wrote already three equations that contain the components of the plane normal. I assume that you know what they are. Now write one more equation that assigns lenght one to that normal. $\endgroup$
    – Kurt G.
    Commented Apr 24, 2022 at 17:45

1 Answer 1

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The problem is, you cannot uniquely solve for $a,b,c$ and $d$, since a single plane can be described by multiple ordered quadruples $(a,b,c,d)$. This is because whenever $k\neq 0$, the plane described by $ax+by+cz=d$ is the same as the one described by $(ka)x+(kb)y+(kc)z=kd$. This agrees with what you have found so far. You have three linearly independent equations, and four variables, so there should be one degree of freedom, represented by $k$.

In order to get a system of equations with a unique solution, you would need to impose a fourth condition. Something like $d=1$ would work.

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