Calculus III: Find the points of the curve... I have to find the points of the curve $$r\left( t \right) =\left( t,{ t }^{ 2 },{ t }^{ 3 } \right) $$ where the osculating plane passes through the point $\left( 2,-\frac { 1 }{ 3 } ,-6 \right)$.
 A: Remember that the osculating plane at $r(t)$ is generated by the vectors $\mathbf{T}(t)$ and $\mathbf{N}(t)$ (tangent and normal). Hence, the binormal vector $\mathbf{B}(t)$ is normal to the osculating plane. Remember that this curve is not parametrized by arc-length, so we have: $$\mathbf{B}(t) = \frac{r'(t) \wedge r''(t)}{\|r'(t) \wedge r''(t)\|}$$ 
You want the values of $t$ that satisfies the equation: $$\left\langle r(t)- \left(2, -\frac{1}{3}, -6 \right), \mathbf{B}(t) \right\rangle = 0$$
Then, to find the points on the curve, compute $r(t)$ for the values of $t$ you found above.

We have $r'(t) = (1,2t,3t^2)$ and $r''(t) = (0,2,6t)$. From this, we get $r'(t) \wedge r''(t) = (6t^2,-6t, 2)$. If the vector $r(t)- \left(2, -\frac{1}{3}, 0 \right)$ is orthogonal to $\mathbf{B}(t)$, it is orthogonal to $r'(t) \wedge r''(t)$ too. We have to solve: $$\begin{align} \left\langle r(t)- \left(2, -\frac{1}{3}, -6 \right), r'(t) \wedge r''(t) \right\rangle = 0\end{align} \\ \langle (t - 2, t^2 + (1/3), t^3 + 6), (6t^2,-6t, 2)\rangle = 0$$
which doesn't seems too hard.
A: Since the osculating plane is made of points of the form:
$$ (t,t^2,t^3)+\lambda_1(1,2t,3t^2)+\lambda_2(0,2,6t) $$
we have to solve, with respect to $t$, the polynomial system:
$$\left\{\begin{array}{rcl} 2 &=& t+\lambda_1 \\ -\frac{1}{3} &=& t^2+2\lambda_1 t+2\lambda_2\\ -6 &=& t^3+3t^2\lambda_1+6t\lambda_2\end{array}\right.$$
that is equivalent to:
$$\left\{\begin{array}{rcl} 2 &=& t+\lambda_1 \\ -\frac{13}{3} &=& 2\lambda_2-\lambda_1^2\\ -14 &=& 3t(2\lambda_2-\lambda_1^2)-\lambda_1^3\end{array}\right.$$
by subtracting from the second equation the square of the first one and by subtracting from the third equation the cube of the first one. By simplifying further:
$$\left\{\begin{array}{rcl} 2 &=& t+\lambda_1 \\ \frac{13}{3} &=& \lambda_1^2-2\lambda_2\\ 14 &=& 13t+\lambda_1^3\end{array}\right.$$
so we have $\lambda_1^3-13\lambda_1+12=0$, from which $\lambda_1\in\{-4,1,3\}$ follows. 
Now the first equation simply gives $t\in\{-1,1,6\}$. Done.
