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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.

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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