Let $L$ be a line that passes through points $a = (1,-1,-2)$ and $b =(2,-1,1)$. Let $V_1$ be the plane $x+y-3z+6=0$.
- Find the equation for $L$.
- Find the equation for the plane $V_2$ that is perpendicular to $V_1$ and contains the line $L$.
I have found $L = (x,y,z)=(b-a)t+a = (1,0,3)t+(1,-1,-2)$.
Thus the parallel vector to $L$ is $ab = (1,0,3)$
The normal vector to the plane $V_1$ can be taken from the coefficients of $x$,$y$ and $z$: $v_1 n = (1,1,-3)$
I now have $ab$ and $v_1 n$ if I find $ab \times v_1n$ I find a vector that is normal to $ab$ and $v_1n$ call this vector $v_2n = (-3,6,1)$.
$v_2n = (-3,6,1)$ is normal to $v_1n$ I tested this with the dot product and must lay in $V_1$ and also be the defining vector for $V_2$ as it is normal to $ab$.
Is my reasoning correct and how do I continue from here?