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CompStang
Sep3-08, 12:58 AM
Find the angle between the surfaces defined by r^2= 9 and x + y + z^2= 1 at the point (2,-2,1)? --- I know this should be extremely simple but it is blowing my mind for some reason. Any help would be greatly appreciated.

HallsofIvy
Sep3-08, 04:26 AM
The angle between two surfaces at a point pf intersection is the angle between their normal vectors at that point. And if you write the surfaces as f(x,y,z)= constant, the normal vector is in the same direction as \nabla f. You probably want to write the first surface in Cartesian coordinates as x2+ y2+ z2= 9. Find the gradients of those two functions and then use the fact that
\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta) where \theta is the angle between the two vectors.

Although this was posted in Precalculus, I don't believe this can be done without using Calculus.