Find the point on the curve $\langle t^3, 3t, t^4 \rangle$ where the plane perpendicular to the curve is parallel to the plane $24x + 6y – 64z = 0$
I assume to find this point I need to parameterize this equation with the curve?
$x \Rightarrow t^3=24u \Rightarrow u = \frac{24}{t^3}$
$y \Rightarrow 3t=6u \Rightarrow u = \frac{t}{2}$
$z \Rightarrow t^4=-64u \Rightarrow u = \frac{-t^4}{64}$
Am I on the right track here?
Best Answer
Hint: The tangent vector at $\bigl(x(t_0),y(t_0),z(t_0)\bigr)$ is normal to the plane $24x-6y+64z=0$.
So you have to solve for $\;\dfrac{3t_0^2}{24}=\dfrac{3}{6}=\dfrac{4t_0^3}{-64}$.