I wish to find a normal vector to the surface given by the interior of the intersection of the cylinder $x^2+y^2=1$ and the plane $x-z=1$. I'm not entirely sure the best way to go about finding a normal vector without parametrizing the surface (which doesn't seem like the best way to go).

I thought to try and use the gradient to find a normal vector but I couldn't figure out exactly what $f(x,y,z)$ I would need to take the gradient of, (and also I would need a point on the surface to find a normal vector at that point).

Any help?

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    $\begingroup$ That intersection is a curve, not a surface. $\endgroup$ – Ivo Terek Mar 13 '16 at 20:21
  • $\begingroup$ I mean the surface defined by the interior of that curve sorry. $\endgroup$ – Samuel Mar 13 '16 at 20:22
  • $\begingroup$ Since that intersection is contained in the plane, the normal vector to the plane, $(1,0,-1)$ will be normal to that surface in every point. $\endgroup$ – Ivo Terek Mar 13 '16 at 20:23
  • $\begingroup$ Could you explain a little more please? $\endgroup$ – Samuel Mar 13 '16 at 20:26
  • $\begingroup$ I wrote an answer, below, trying to explain more. $\endgroup$ – Ivo Terek Mar 13 '16 at 20:31

Since that surface is contained in the plane $x-z=1$, which has as a possible normal vector $(1,0,-1)$ (because every normal vector to a plane given by $ax+by+cz=d$ is a multiple of $(a,b,c)$), then this same vector is normal to the surface. Indeed, if you take any tangent vector to the surface, it is tangent to the plane, and hence orthogonal to $(1,0,-1)$.

The cylinder is not important here, in this aspect. The crucial issue is that the surface is contained in a plane.


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