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)*(1,1,-3) = (-3,12,3) \\e (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`.