Find the point on the curve where the tangent line is orthogonal to the plane. Find the point on the curve $$ \overrightarrow{\rm r}(t) = \langle t^3+3t, t^2+1, \ln (1+2t) \rangle  ,   0\leq t \leq \pi$$ where the tangent line is orthogonal to the plane $$15x+4y+0.4z=10$$. 
So I was able to find the derivative: $$\overrightarrow{\rm r}'(t) = (3t^2+3)\overrightarrow{\rm i} + 2t\overrightarrow{\rm j} +(2\div(1+2t))\overrightarrow{\rm k} $$
and the normal vector to the plane: $$\overrightarrow{\rm n}=\langle 15,4,0.4\rangle$$
I attempted to solve by finding the dot product of the $\overrightarrow{\rm r}'(t)$ and the $\overrightarrow{\rm n}$ and I got $ t=-0.507614$
This doesn't work out when I plug it back in. I'm not sure what I'm doing wrong? 
 A: I’m assuming that the way you solved for $t$ was by setting $\vec r'\cdot\vec n=0$. This equation says that $\vec r'$ is orthogonal to $\vec n$, but that makes it parallel to a plane with normal $\vec n$, not orthogonal to it. What you want is for $\vec r'$ to be parallel to $\vec n$, that is $\vec r'=\lambda \vec n$ for some nonzero scalar $\lambda$. Since you’re working in $\mathbb R^3$, you can avoid introducing an extra variable by using the equivalent condition $\vec r'\times\vec n=0$. This will give you a system of three equations in $t$ to solve.
A: We have that
$$\overrightarrow{\rm r}'(t) = (3t^2+3)\overrightarrow{\rm i} + 2t\overrightarrow{\rm j} +(\color{red}2\div(1+2t))\overrightarrow{\rm k}$$
As noticed by amd in the comments, of course we need to find t by the condition
$$\vec r'(t) \times \vec n=\vec 0$$
A: The derivative should be $\overrightarrow{\rm r}'(t) = (3t^2+3)\overrightarrow{\rm i} + 2t\overrightarrow{\rm j} +(2\div(1+2t))\overrightarrow{\rm k}\\$
. The kth direction should have a 2 in numerator. Now all you do is equate the norm to this derivative to ensure they have the same direction (then tangent is perpendicular to plane) should give you $t=2$ then plug this in for your equation of the plane.
A: $r = t^3+3t,t^2+1,\ln(1+2t)\\
\frac {dr}{dt} = 3t^2 + 3, 2t, \frac {2}{2t + 1}$
If $\frac {dr}{dt}$ is normal to your plane. 
$\frac {dr}{dt}$ points in the same direction as $(15,4,0.4)$
or for some $\lambda, \frac {dr}{dt} = \lambda (15,4,0.4)$
At this point, by inspection we see that at $t = 2$
$\frac {dr}{dt} = (15,4,0.4)$
If we were not so lucky, and we wanted to use the dot product.
then $\frac {dr}{dt}\cdot (15,4,0.4) = \|\frac {dr}{dt}\|\|(15,4,0.4)|$
but $\|\frac {dr}{dt}\| = \sqrt{9t^4 + 22t^2 + 9 + \frac {4}{4t^2 + 4t + 1}}$
looks like a mess to work with.
If it were something nicer, I am still not sure I would use that formula.
I might say:
$T = \frac {\frac {dr}{dt}}{\|\frac {dr}{dt}\|}$ is the unit tangent vector.
$\frac {dT}{dt}$ is normal to $T.$  and hence in your plane when T takes on the correct direction and 
$\frac {dT}{dt} \cdot (15,4, 0.4) = 0$
