Hello, mathstresser!
What is the equation of the plane that contains the line \(\displaystyle \,\begin{array}{ccc}x\:=\:3\,+\,2t \\ y\:=\:t \\ z\:=\:8\,-\,t\end{array}\)
and is parallel to the plane \(\displaystyle 2x\,+\,4y\,+\,8z\:=\:17\)
To write the equation of a plane, we need a point \(\displaystyle (x_1,y_1,z_1)\)
\(\displaystyle \;\;\)and the normal vector of the plane: \(\displaystyle \,\vec{n}\,=\,\langle a,b,c\rangle\)
Then the equation is: \(\displaystyle \,a(x\,-\,x_1)\,+\,b(y\,-\,y_1)\,+\,c(z\,-\,z_1)\;=\;0\)
The plane we want contains that line so it contains
any point on the line.
Let \(\displaystyle t\,=\,0\) and we have a point: \(\displaystyle \,(x_1,y_1,z_1)\,=\,(3,0,8)\)
The given plane has normal direction: \(\displaystyle \,\vec{n}\,=\,\langle2,4,8\rangle\)
The plane we want is parallel to it, so it has the
same normal vector.
Can you finish it now?
