Let $C$ be the Intersection parametric equation that occurs from intersecting with the plane $x-z-1=0$ and the cylinder $x^2+y^2=8$. on that parametric equation there is a point $(2,2,1)$. let $L$ be the line Tangent to the parametric equation in the given point. find the point that is on that Tangent.
I used the cylinder coordinates and got that $x=8cos\theta$ , $y=8sin\theta$ , $z=z$ I found $z$ from the plane equation , I did $8cos\theta-z=1$ so I got $z=8cos\theta-1$ from that we get the Intersection parametric equation $r(t)=(8cos\theta,8sin\theta,8cos\theta-1$ and then to find the tangent I just did a derivative and got $r'(t)=(-8sin\theta,8cos\theta,-8sin\theta)$
and here I got stuck , I don't know how to continue.. Would appreciate any helps any tips it is more important than the final answer and sorry for the english mistakes I tried to translate hope it is understandable
Equation of cylinder is $x^2 + y^2 = 8 = (2\sqrt2)^2$
The intersection curve of the cylinder and the plane $x-z = 1$ can be parametrized as,
$r(t) = (2 \sqrt2 \cos t, 2\sqrt2 \sin t, 2 \sqrt2 \cos t - 1)$
So $r'(t) = (-2\sqrt2 \sin t, 2\sqrt2 \cos t, - 2\sqrt2 \sin t)$
Please note that at point $(2, 2, 1)$ on intersection curve, $t = \frac{\pi}{4}$ and $r'(\frac{\pi}{4}) = (- 2, 2, - 2)$
So parametric equation of tangent line at $(2, 2, 1)$ can be written as $(2, 2, 1) + s (-2, 2, -2)$.