Determine the equation of a plane orthogonal to the plane: $ (P) : 3x-y+3z-2=0$ and whose intersection with $(P)$ is a line from the $xOy$ plane.
My idea is the following: since the planes are orthogonal it means that the dot product of the normal vectors is $0$.I think that since the line is from $xOy$, the dot product would be $3a-b=0$, but I'm not sure. Could you please show me what a line from the $xOy$ plane looks like?
Let us first see what's the line which belongs both to $P$ and to the plane $xO$. For that we solve the system$$\left\{\begin{array}{l}3x-y+3z=2\\z=0,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}3x-y=2\\z=0\end{array}\right.$$Two solutions of this system are $(1,1,0)$ and $(0,-2,0)$. So, you are after a plane which $ax+by+cz=d$ which contains these two points and which is orthogonal to $P$. So, you solve the system$$\left\{\begin{array}{l}a+b=d\\-2b=d\\3a-b+3c=0.\end{array}\right.$$One solution is $a=\frac32$, $b=-\frac12$, $c=-\frac53$, and $d=1$. So, take the plane$$\frac32x-\frac12y-\frac53z=1\left(\iff9x-3y-10z=6\right).$$
