Vector Valued Functions, Find some value at point Suppose that $r$ is a vector valued function of $t$. Now, $r_0=\langle 2,2,2\rangle$ and $r_1$ is in the $y,z$ plane. If $r' \times \langle 2,3,4\rangle=0 \forall t$, how can I find what $r_1$ is?
I thought that I use matrices to find $r'$, but I can't figure out how to do that, or if that even works. 
 A: Hint 1: $a\times b = 0$ implies $a$ and $b$ are linearly dependent.
Hint 2': Use the mean value theorem, there exists some $0 < c < 1$ such that
$$ r(1) - r(0) = r'(c). $$
It is less restrictive than the fundamental theorem.
Hint 3: Exploit $r_z(1) = 0$.
Solution: 
Since $r'(c)$ and $\langle 1,2,3 \rangle$ are linearly dependent, there is some $\alpha\in\mathbb R$ such $r'(c) = \alpha \langle 1,2,3 \rangle$. From $r_z(1) = 0$ and $r_z(1) - r_z(0) = \alpha 3$ we conclude that $\alpha = \frac13$ and thus $r(1) = \langle \frac23, \frac13, 0 \rangle$.
A: \begin{align}
& (r(1)-r(0))\times (1,2,3) = \int_0^1 r'(t)\,dt\times(1,2,3) \\[10pt]
= {} & \int_0^1 r'(t)\times(1,2,3)\,dt = \int_0^1 (0,0,0)\,dt = (0,0,0).
\end{align}
So
$$
((x,y,0)-(1,1,1))\times(1,2,3) = (0,0,0).
$$
If you now evaluate the cross-product on the left, you get two equations in two variables.
A: Maybe it will be easier to picture if you see the problem physically. Suppose the function describes the motion of a particle; the particle starts at $(1, 1, 1)$ and is moving in the direction of the vector $(1, 2, 3)$ at all times (because its velocity is parallel to this direction) although we don't know what speed it's moving at. That is, the particle is traveling in a straight line.
We're told that at some point, the $z$-coordinate of the particle is $0$. We have
$$
(1, 1, 1) + (s, 2s, 3s) = (?, ?, 0)
$$
Here $s$ is some possibly very complicated function of the time $t$ that we don't know, but it doesn't matter. The third coordinate tells us $s = -1/3$ at the point in question, and so we get $(2/3, 1/3, 0) $ as desired.
