Respuesta :
Let [tex]L[/tex] be the line given by the vector equation
[tex](-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}[/tex].
First, we use the director vectors of the lines L1 and L2 to get the
vector equation of the plane containing them, which we denote by [tex]\Pi_1[/tex]. This is,
[tex]\\\\\Pi_1 : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\[/tex]
We observe that [tex]\vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0)[/tex]. Therefore, the vector equation of [tex]\Pi_1[/tex] defines a plane and [tex]\vec{N}[/tex] is a normal vector to [tex]\Pi_1.[/tex]
Finally, the vector equation for the wanted plane, which we denote by [tex]\Pi[/tex], is
[tex]\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .[/tex]
Thus, if [tex]s = 0[/tex], then [tex]L \subset \Pi[/tex] and since [tex]\vec{N}[/tex] is parallel to [tex]\Pi[/tex], then it is perpendicular to [tex]\Pi_1[/tex].
