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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 $3x+y+z+2=0$?

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up vote 3 down vote accepted


  1. The tangent line at $t$ has direction $\alpha'(t)$
  2. The normal of a plane given by $ax + by + cz + d = 0$ is $(a,b,c)$.
  3. 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 $(3-3t^2,6t, 3 + 3t^2) \cdot (3,1,1) = 0$.


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

And we get

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

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But ... it was homework. – Gigili Jun 23 '12 at 2:47
@Gigili I am getting a bit carried away, aren't I? I've fixed it somewhat. I can't make the spoiler thing work. – talmid Jun 23 '12 at 2:55
Forget about the spoiler. It's the same as in Mohamed's answer anyway. – talmid Jun 23 '12 at 3:07
Weird, I've spent about five minutes and it doesn't work! Silly thing. +1 anyway, and thanks for editing your answer. – Gigili Jun 23 '12 at 3:21
Thank you for your answer. I initially thought about approach similar to this but at the end I wasn't sure. – MasterM Jun 23 '12 at 10:57

$n=(3,1,2)$ 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 $t=2$

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