Let $C$ be the two-sided cone $z^2=x^2+y^2$
1) Let $S$ be the set of all points ($x,y,z$) in $C$ such that the normal line of $C$ at ($x,y,z$) is perpendicular to the vector $v= \hat i+\hat j+\sqrt{2}\hat k$. Prove that $S$ lies on a line $L$.
My resolution: The normal line to the surface $C$ is parallel to the gradient of $f$. Therefore a vector representation of the normal line is $\vec u=(-2x)\hat i+(-2y)\hat j+(2z)\hat k$. We want $\vec u\cdot\vec v=0$ which leaves us with the surface equation: $x+y-\sqrt{2}z=0$ Doesn't a surface lie on an infinite number of lines?