a sphere has equation: x^2+y^2+z^2=2 find the equation of the tangent plane at (0,1,1) on the sphere
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Hi The tangent plane at a point of a sphere is perpendicular to where is the centre of the sphere. This gives you a vector which is normal to the plane. As we also have that is a point of the plane, you can determine the equation you're looking for...
Originally Posted by szpengchao a sphere has equation: x^2+y^2+z^2=2 find the equation of the tangent plane at (0,1,1) on the sphere If then the normal vector of the tangent plane can be calculated by: With your question you get: Plug in the coordinates of the tangent point and the normal vector is: Since the point T(0, 1, 1) is located on the plane the equation of the plane becomes:
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