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 $x=0$; $z + y \tan\alpha = 0$.


closed as off-topic by Jean Marie, Michael Hoppe, Ken Duna, user26857, John B Apr 6 '17 at 20:50

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  • $\begingroup$ No personal thoughts ? "I give you the exercice text and you have to produce me an answer"...What have tried ? What is the blocking point ? I vote to close this question unless you say something personal. $\endgroup$ – Jean Marie Apr 6 '17 at 13:15
  • 1
    $\begingroup$ First I find the equation of normal planes to the given curve. but i dont know exactly what i have to do next $\endgroup$ – Hitman Apr 6 '17 at 13:16
  • $\begingroup$ first i thought that the equation of normal planes should satisfy the equation of given line but it is not working $\endgroup$ – Hitman Apr 6 '17 at 13:17
  • $\begingroup$ All right, then write this equation for us : we will be able to say you how to go ahead. $\endgroup$ – Jean Marie Apr 6 '17 at 13:18
  • $\begingroup$ -a sint (x-x_o) + a sinα cost (y-y_o) + a cosα cost (z-z_o) =0 $\endgroup$ – Hitman Apr 6 '17 at 13:23

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.

  • $\begingroup$ I do not understand how you find the value of h. $\endgroup$ – Hitman Apr 6 '17 at 14:07
  • $\begingroup$ I express the fact that $(x,y,z)=(x_0,y_0,z_0)$ given by (2) must verify equation (1), giving finaly (3). $\endgroup$ – Jean Marie Apr 6 '17 at 14:10

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