Vector Calculus Question- Planes and Curves

Question

Will you please help me in the following?

Let $\pi$ be a plane perpendicular to the curve: $$ \gamma(t) = (5\cos t, 5\sin t,-2t) $$ at the point $(x(t_0), y(t_0 ) ,z(t_0)) $ .

We also know the distance from $(0,0,0)$ to the plane $\pi$ is 7.

What is the value of $t_0$?

All I know is that the normal to the plane at $t_0$ is $(-5\sin t_0, 5\cos t_0,-2) $ and that substituting into the formula for distance from a point to a plane doesn't help (since I don't have the free term of the planes formula)

Will you please help me?

Thanks

Answer 1

Your plane at a time $t$ is $N_t:=\{(x,y,z)\in\mathbb{R}^3:-5\sin t(x-5\cos t)+5\cos t(y-5\sin t)-2(z-2t)=0\}$

Thanks to Lagrange's method you can find $(x(t),y(t),z(t))\in N_t$ minimizing $\|v\|^2:v\in N_t$.

Then you can solve $x^2(t)+y^2(t)+z^2(t)=49$ and get $t_0$

I don't know if it's the optimal method, though, because it requires a lot of calculations. But it's a method.