Given the curve $c(t)=\langle sint+1,2-cost,sint+1\rangle$ find cylinder and plane surface equation.
The cylinder is easy: $$ \cos t=2-y\\\sin t=x-1 $$ then cylinder equation is $(x-1)^2+(2-y)^2=1$. But what about the plane?
It looks like $z=x$ so is it correct that the plane is $z=x$?
The point $(1,1,1)$ is on the curve, $ t= 0$.
Equation of a plane passing through $ (1,1,1)$ with normal vector $\vec n$:
$n \cdot ( \vec r - (1,1,1)) = 0$.
Every point $\vec r(t)$ of the curve
lies in the plane:
$\vec n \dot ( \sin(t), 1- \cos(t), \sin(t)) = 0$.
With $\vec n := (a,b,c)$ we find:
$(a + c) \sin(t) + b (1- \cos(t)) = 0$ ,
$0 \le t \lt 2π.$
$(a+c) \sin(t) -b\cos(t) = - b$.
$t =π/2$: $a + c = - b$ ;
$t = π$ : $b = -b,$ hence $b= 0$.
$\vec n = a(1,0,-1)$.
Plane:
$(1,0,-1)( (x,y,z) - (1,1,1)) = 0$;
$(x-1) - (z-1) = 0$;
$x- z = 0$.
Makes sense?