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Advanced Math/Liear algebra finding the equation of a plane tangent to a sphere

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Question
I am confused by what to do for this problem.
Consider the sphere S of radius 5 centred at (1, 0,2).   Find the equation of the plane tangent to S at the point (5,-3,2).

I have attempted this so far:

normal vector: (5,-3,2)-(1,0,2)= <-4,3,0>

ax+by+cz=d     Plug point (5,-3,2) to find d
d= -29
the equation of the plane is thus -4x+3y=-29

Is this how you do it/the right answer? I have no idea. Thanks for any help.

Answer
Good job. Key insight was the fact that the vector normal to the tangent plane is the vector from the sphere's center to the point on its boundary. My only tweak would be the minus signs in the normal vector (i.e., 5-1 = +4).

The scalar equation you derived for the plane pops out of the definition for the locations of points in a plane, i.e., points contained in all vectors perpendicular to a given vector

n・(P-P0) = (4,-3,0)・(x-5,y+3,z) = 4(x-5) - 3(y-3) = 0   <- zero means perpendicular.

Multiplying this out gives your equation, including d = -29.

Very glad you showed your work.

Randy

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randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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