Question

Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the two lines
L1: x=2t+1,y=-t-1,z=6t+5

L2:x=t+1,y=t-1,z=-3t+5

Answer 1

Let $L$ be the line given by the vector equation

$(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}$.

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 $\Pi_1$. This is,

$\\\\\Pi_1 : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}$

We observe that $\vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0)$. Therefore, the vector equation of $\Pi_1$ defines a plane and $\vec{N}$ is a normal vector to $\Pi_1.$

 

Finally, the vector equation for the wanted plane, which we denote by $\Pi$, is

$\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .$

Thus, if $s = 0$, then $L \subset \Pi$ and since $\vec{N}$ is parallel to $\Pi$, then it is perpendicular to $\Pi_1$.