Proof about parametrized curve

[throwaway7050]

"Suppose phi(t) is a parametrized curve in R³ such that the 3rd derivative of phi(t) is identically 0. Prove there exists a plane P, which is a subset of R^3, such that for all t, phi(t) is in P."
The following are some of my thoughts: Since the 3rd derivative is identically 0, then we know that the 2nd derivative is constant i.e. of the form (a_1, a_2, a_3). Similarly, we can conclude that the 1st derivative is linear i.e. of the form (a_1*t + b_1, a_2t* + b_2, a_3*t + b_3). We integrate again to see that each of the components in the original function phi(t) is quadratic.
I'm stuck on where to go from here. Any help or insight would be greatly appreciated! Thanks!

 

[daon2]

"Suppose phi(t) is a parametrized curve in R³ such that the 3rd derivative of phi(t) is identically 0. Prove there exists a plane P, which is a subset of R^3, such that for all t, phi(t) is in P."
The following are some of my thoughts: Since the 3rd derivative is identically 0, then we know that the 2nd derivative is constant i.e. of the form (a_1, a_2, a_3). Similarly, we can conclude that the 1st derivative is linear i.e. of the form (a_1*t + b_1, a_2t* + b_2, a_3*t + b_3). We integrate again to see that each of the components in the original function phi(t) is quadratic.
I'm stuck on where to go from here. Any help or insight would be greatly appreciated! Thanks!

How I would approach it (not necessarily the best way):

So then \(\displaystyle \phi(t) = Pt^2+Qt+R\) for points \(\displaystyle P,Q,R\in \mathbb{R}^3\).

Three points on the curve are: \(\displaystyle (1) R, (2) P+Q+R, (3) P-Q+R\). We can find vectors \(\displaystyle \overrightarrow{a} = (2)-(3) = \overrightarrow{2Q}, \overrightarrow{b} = (2)-(1) = \overrightarrow{P+R}\) which are in the plane containing these points. A normal vector to this plane is \(\displaystyle \overrightarrow n = \overrightarrow a \times \overrightarrow b = \langle n_0, n_1, n_2\rangle\). Assuming \(\displaystyle R=(x_0,y_0,z_0)\) (choosing an arbitrary point in the plane)then the plane is given by

\(\displaystyle \overrightarrow{n} \cdot (\langle x,y,z\rangle-R) = n_0(x-x_0) + n_1(y-y_0)+n_2(z-z_0)=0\).

Now let \(\displaystyle \phi(z) = Pz^2+Qz+R\) be an arbitrary point on the curve, show that \(\displaystyle \overrightarrow{n} \cdot (\phi(z)-R)=0\)