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

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

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Hoppe, Ken Duna, user26857, John B
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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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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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