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Find the equation of a normal plane at a point (1,2,3) on the intersecting line of $ x^2+y^2+z^2=14 $ and $x^2+y^2=5$.


Is the definition of the normal plane here the plane contains two normal vector at the point?

Is is okay if I find the cross product of two tangent vector at the point,
and use it to find an equation of plane which I think is the mentioned normal plane?

It results in x+2y=5.

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  • $\begingroup$ As stated the problem is not so clear. By the solution given it seems that we are required to find a tangent plane to the intersection line at that point. $\endgroup$
    – user
    Jul 27 '18 at 9:20
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It seems that the problem is to find the tangent plane to the intersection line at that point. Therefore we need to find the normal vector to that line, that is normal vector to the plane.

Since the intersection is the circle

  • $x^2+y^2=5$

  • $z=3$

and the tangent vector at that point is $t=(2,-1,0)$, the normal vector is $n=(1,2,0)$ and the result follows.

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  • $\begingroup$ I've written every information in the problem. Do you think this is somehow wrong? $\endgroup$
    – NK Yu
    Jul 27 '18 at 12:08
  • $\begingroup$ and I assumed the normal plane to have two normal vector that each came from the first and second equation. Is it not correct? $\endgroup$
    – NK Yu
    Jul 27 '18 at 12:09
  • $\begingroup$ @NKYu The definition of normal plane is not completely clear to me in this case. $\endgroup$
    – user
    Jul 27 '18 at 12:10
  • $\begingroup$ @NKYu The intersection of the given surfaces at that point is a circle and the plane is a tangent plane to the circle at that point. $\endgroup$
    – user
    Jul 27 '18 at 12:11

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