Prove that the Space Curve lies entirely in a single plane and find an equation for it

Question

I am trying to prove that the space curve $$r(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \\ \end{bmatrix} = \begin{bmatrix} 2t - \cos(t) +3 \\ \sin^2(t) +4t \\ 1/2 (-\cos^2(t)+2\cos(t) +1)\\ \end{bmatrix} $$

lies entirely in a single plane and find that plane's equation.

I tried picking 3 points

$$r(0) = P(1, 0, 1)$$

$$r(\pi) = Q(2\pi+1, 4\pi, -1)$$

$$r(2\pi) = R(4\pi-1, 8\pi, 1).$$

I picked $r_0 = P(1, 0, 1)$, and calculated ${PQ}\times{PR}$ to find a normal vector $\vec{n}$ to the plane. I found $\vec{n} = <16\pi, 8\pi - 4, 0>$, and then attempted to dot $\vec{n}$ with $\vec{v(t)}$. My logic was if the curve lies in a plane, this dot product has to equal $0$ as they would always be orthogonal to each other. However my dot product did not equal $0$ and I don't know where to go from here.

Answer 1

An elegant way to see that the curve is contained on a plane is to look for linear combinations of its coordinates which are constant.

In that case, I see that $x(t)-1/2y(t) +z(t) = 3$, which shows that the curve is contained in the plane $x-y/2+z=3$.

Answer 2

You have done some simple math errors. $$r(0)=P(2,0,1)\\r(\pi)=Q(2\pi+4,4\pi,-1)\\r(2\pi)=R(4\pi+2,8\pi,1)$$ With these I get $$\vec n=8\pi(2,-1,2)$$ The velocity is $$\vec v=(2+\sin t,2\sin t\cos t+4, \cos t\sin t-\sin t)$$ Then $$\vec n\cdot \vec v=8\pi(4+2\sin t-2\sin t\cos t-4+2\sin t\cos t-2\sin t)=0$$