A line L is given as (9/4,-1/4,0) + $\lambda$(1,3,8). A plane $\pi$ is parallel to both the line L and the line (4,6,2) + $\Phi$(3,-2,0). Given that plane $\pi$ contains the point (1,2,0). Find the Cartesian equation of a plane.
I cant get around the question because the direction vectors of both lines are parallel to the plane and doesn't help me in finding the normal vector of the plane in any way.
Best Answer
Hint
If the plane is parallel to the lines given by $\vec x = \vec p + \lambda \color{blue}{\vec v}$ and $\vec x = \vec q + \lambda\color{blue}{ \vec w}$, then $\vec n = \color{blue}{\vec v} \times \color{blue}{\vec w}$ is a normal vector of the plane. A Cartesian equation of the plane with normal vector $\vec n = (a,b,c)$ and containing a point $(x_0,y_0,z_0)$ is given by: $$a\left( x-x_0 \right)+b\left( y-y_0 \right)+c\left( z-z_0 \right)=0$$