Point on a curve where the normal plane is parallel to another plane A particle follows the curve $r ( t ) = ( \frac{4}{t + 1}, t^3 , t^4 + 1 )$ for time $t \ge 0$
Find the point $P ( x,y,z )$ on the curve such that the normal plane to the curve at P is parallel to the plane: $\frac{-1}{9} x + 3 y + 8 z = 0$
I have no idea where to start, all I know is $(-1/9,3,8)$ would be a line orthogonal to the plane and the normal plane to the curve at $P$.
The answer is $(4/3,8,17)$ so I'm guessing they somehow found $t=2$ and plugged that in to r(t), but can someone explain to me the logical steps to take to solve this problem.
 A: The derivative of the curve at $t$ is $v=(-4/(t+1)^2,3t^2,4t^3).$ Then you need to set an undetermined scalar multiple $kv$ to the vector $(-1/9,3,8)$ normal to the plane. This gives from the second coordinate that $3kt^2=3,$ so we know $kt^2=1.$ Then from the third coordinate we get $4kt^3=8,$ which means using $kt^2=1$ that $4kt^3=4(kt^2)t=4\cdot 1 \cdot t=8, $ so $t=2$ is forced. Also since $kt^2=1$ we have $k \cdot 2^2=1$ and then $k=1/4.$ Checking at the first coordinate we have $(1/4)\cdot (-4/(9))=-1/9$ as it should be.
A: The given plane has the equation $u \cdot n = 0$ with normal vector $n = (-1/9, 3, 8)$.
The curve is $r(t) = (4/(t+1),t^3, t^4+1)$.
The tangent unit vector $T$ of the curve is
$$
T 
= 
\frac{dr}{ds} = \frac{\dot{r}}{\lVert \dot{r} \rVert} 
= 
\left(
-\frac{1}{(t+1)^2}, 3 t^2, 4 t^3
\right) 
/
\lVert \dot{r} \rVert
$$ 
As fellow user coffeemath has probably realized before me, $T$ is orthogonal to the normal plane and thus pointing in the same direction as $n$ (or its negative), so we just take $\dot{r}$ (orthogonal is good enough, normality is not needed) and find out when it turns into a scalar multiple of $n$, for $t \ge 0$:
$$
\dot{r}(t) = \alpha n \wedge t \ge 0 \iff
(-1/9,3,8) = \alpha \left(
-\frac{1}{(t+1)^2}, 3 t^2, 4 t^3
\right) \wedge t \ge 0 \iff t = 2 \wedge \alpha = 1 
$$
and this is
$$
r(2) = (4/3,8,17).$$
Here is a visualization with tangent vectors for $t \in \{ 0, 1, 3/2, 2 \}$. The normal of the plane is attached to the origin.

A: (Take any two vectors in the given plane A and B. 
By differentiation find a tangent T to the curve.
The scalar triple product (A X B . T)  = 0, so find $t.$
