- #1

- 129

- 0

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