It is proved that on the smooth surface f (x, y, z) = 0, the normal of the point closest to the origin must pass through the origin

It is proved that on the smooth surface f (x, y, z) = 0, the normal of the point closest to the origin must pass through the origin

First of all, if the surface passes through the origin, the point closest to the origin on the surface is of course the origin, so the normal of the surface at the origin passes through the origin. Let's just prove that the surface can't pass through the origin. Set point (x, y, z) ≠ (0,0,0) to minimize the distance from the point to the origin, that is, to minimize x ^ 2 + y ^ 2 + Z ^ 2, because the point (x, y, z) is required to be on the surface, According to the Lagrange multiplier method, the constructor f (x, y, Z, λ) = x ^ 2 + y ^ 2 + Z ^ 2 - λ f (x, y, z) is used to calculate the partial derivative of X and make it equal to 0, which has 2x - λ f'x = 0. Similarly, we can get x = λ f'x / 2, y = λ f'y / 2, z = λ f'z / 2 (there is no need to calculate the specific value of λ here). Write the passing point of the surface (λ f'x / 2, λ f'y / 2, The normal equation of (x - λ f'x / 2) / f'x = (Y - λ f'y / 2) / f'y = (Z - λ f'z / 2) / f'z can be verified by substituting x = y = z = 0 into the above equation, so the normal passes through the origin