Normal planes to a curve passing through a line

Question

Show that the normal planes to the curve

$$x=a \cos t$$ $$y=a \sin\alpha \sin t$$ $$z=a \cos\alpha \sin t$$

are passing through the straight line $x=0$; $z + y \tan\alpha = 0$.

Answer 1

As I see now that you have worked on the subject, here is a solution:

The normal plane at epoch $t_0$ is the plane having as normal vector a tangent vector to the curve,

• i.e., taking the derivatives, with equation

$$\tag{1}x(-a \sin t_0)+y(a \sin \alpha \cos t_0)+z(a \cos \alpha \cos t_0)=h.$$

• with $h$ determined by the fact that this plane passes through point

$$\tag{2}(x_0=a\cos t_0 , y_0=a \sin \alpha \sin t_0 , z_0=a \cos \alpha \sin t_0)$$

which means that, replacing in (1) $(x,y,z)$ with $(x_0,y_0,z_0)$ given by (2), we get:

$$\tag{3}h= \left[-1+(\sin \alpha)^2 +(\cos \alpha)^2\right)]a^2\cos t_0\sin t_0=0.$$

Thus this is a plane passing through the origin.

It remains to show that $(x,y,z) \in $ (L) (the line) $\implies (x,y,z) \in $ (P) the plane (recall that implication between characteristic properties corresponds to set inclusion), i.e.,

$$\begin{cases}x=0\\ z = - y \tan \alpha \end{cases} \ \ \ \implies x(-a \sin t_0)+y(a \sin \alpha \cos t_0)+z(a \cos \alpha \cos t_0)=0.$$

which is almost immediately established.