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.
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.