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say we have a curve f(x,y,z)

and the tangent plane to the curve at a point (x_{o},y_{o},z_{o}) is given by

[tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(x-x_{o}) + [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) + [tex]\frac{{\partial d}}{{\partial z}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0

then would the normal plane to the curve be either of the following?

[tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial z}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0

[tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0

[tex]\frac{{\partial d}}{{\partial z}}[/tex](x_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0

[tex]\frac{{\partial d}}{{\partial z}}[/tex](x_{o},y_{o},z_{o})(x-x_{o}) - [tex]\frac{{\partial d}}{{\partial y}}[/tex](x_{o},y_{o},z_{o})(y-y_{o}) - [tex]\frac{{\partial d}}{{\partial x}}[/tex](x_{o},y_{o},z_{o})(z-z_{o}) = 0

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# The equation of a normal plane to a curve - confusion

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