Find the equation in the general form of the sphere described

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To find the equation of the sphere tangent to the plane x - 3y + 4z + 23 = 0 at the point (1, 4, -3) with a radius of √26, the normal vector to the plane is crucial as it indicates the direction toward the sphere's center. The center of the sphere can be determined by moving along the normal vector from the tangent point by the radius distance. Since the normal vector has two possible directions, there are two potential centers for the sphere. The general form of the sphere's equation can then be derived using the center coordinates and the given radius. This approach allows for the correct formulation of the sphere's equation in relation to the specified plane and point.
lolphysics3
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Homework Statement


tangent to x-3y+4z+23=0 at (1,4,-3) with radius sqrt(26)


Homework Equations





The Attempt at a Solution



well i initially thought it's possible to set up a system of equations but it didn't work out.
 
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lolphysics3 said:

Homework Statement


tangent to x-3y+4z+23=0 at (1,4,-3) with radius sqrt(26)


Homework Equations





The Attempt at a Solution



well i initially thought it's possible to set up a system of equations but it didn't work out.
Can you find a normal to the plane? It will point toward the center of the sphere (there will be two of them, though). And you know the radius of the sphere, so you should be able to find the center of the sphere.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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