At which points tangent to a curve is parallel to given plane? At which points on the curve $\alpha(t):=(3t-t^3,3t^2,3t+t^3)$ the corresponding tangent lines are parallel to the plane $3x+y+z+2=0$?
 A: Hints:


*

*The tangent line at $t$ has direction $\alpha'(t)$

*The normal of a plane given by $ax + by + cz + d = 0$ is $(a,b,c)$.

*In order for a line to be parallel to a plane, its direction vector must be orthogonal to the normal to the plane.


This will give you a polynomial equation in $t$.


 The tangent line at $t$ has slope $\alpha'(t) = (3-3t^2,6t, 3 + 3t^2)$, and for it to be parallel to the plane given by the equation $3x+y+z +2= 0$, with normal $(3,1,1)$, we must have $(3-3t^2,6t, 3 + 3t^2) \cdot (3,1,1) = 0$.

Then

 $\begin{align*}&(3-3t^2,6t, 3 + 3t^2)\cdot (3,1,1)=9-9t^2+6t+3+3t^2=-6t^2+6t+ 12=0\\ &\Leftrightarrow t^2-t-2=0 \Leftrightarrow(t+1)(t-2)=0\end{align*}$

And we get

 $t = -1$ or $t=2$.

A: $n=(3,1,2)$ is a normal vecttor to the plane . We have : $\alpha'(t)=(3-3t^2,6t,3+3t^2)$ is a tangent line director vector is parallel to the  plane  if  it's  normal to $n$, thus : $n.\alpha'(t)=0$, thus : $$3(1-t^2) + 2t + 1+t^2=0$$ $$-2t^2 + 2t +4 =0 $$   $$t^2-t-2 =0 $$ gives : $t=-1$  or  $t=2$
