# Find cartesian equation of the plane that is parallel to the two lines and passes a point

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,

• Please use formulas to show exactly what you did. Commented Apr 24, 2022 at 16:47
• I've edited my question Commented Apr 24, 2022 at 17:01
• 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. Commented Apr 24, 2022 at 17:12
• can you please write the 4'th equation? Commented Apr 24, 2022 at 17:39
• 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. Commented Apr 24, 2022 at 17:45

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.