Angle Between two surfaces and a point

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SUMMARY

The discussion focuses on calculating the angle between two surfaces defined by the equations r² = 9 and x + y + z² = 1 at the point (2, -2, 1). The angle is determined by the normal vectors at the intersection point, which can be found using the gradients of the surfaces. The first surface is expressed in Cartesian coordinates as x² + y² + z² = 9. The relationship between the angle θ and the normal vectors is established through the equation u·v = |u||v|cos(θ).

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  • Understanding of gradient vectors and their significance in multivariable calculus
  • Familiarity with surface equations in Cartesian coordinates
  • Knowledge of vector operations, specifically dot products
  • Basic concepts of angles between vectors in three-dimensional space
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  • Study the calculation of gradients for multivariable functions
  • Learn about the properties of normal vectors in relation to surfaces
  • Explore the application of the dot product in finding angles between vectors
  • Review the fundamentals of multivariable calculus, focusing on surface intersections
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Students and professionals in mathematics, particularly those studying calculus and vector analysis, as well as anyone interested in understanding the geometric relationships between surfaces in three-dimensional space.

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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.
 
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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 [itex]\nabla f[/itex]. 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
[itex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/itex] where [itex]\theta[/itex] is the angle between the two vectors.

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