Question

Find a unit vector normal to the surface x^3+y^3+3xyz at the point (1,2,-1).

Answer 1

Hi gradient of a scalar surface gives you the vector normal to it.
Please find the attachment.
Hope this helps.

Answer 2

Given:

Equation of surface is f(x,y,z)=$3x^3+3y^3+3xyz$

To Find:

Find a unit vector normal to the surface.

Step-by-step explanation:

the gradient of  f(x,y,z) at point x,y,z is vector normal to the surface at this point.

         $f(x,y,z)=x^3+y^3+3xyz\\gradiant of f=(\frac{id}{dx} +\frac{jd}{dy} +\frac{kd}{dz} )(x^3+y^3+3xyz)\\=i(3x^2+3yz)+j(3y^2+3xz)+k(3xy)\\\\\textrm{at point} (1,2,-1)\\=-3i+9j+6k$

Now a unit vector normal is given by

  =$=\frac{-3i+9j+6k}{\sqrt{3^2+9^2+6^2} } =\frac{-i+3j+2k}{\sqrt{14} }$

Hence,unit vector normal is given by$\bf \frac{-i+3j+2k}{\sqrt{14} }$