How do you find a unit vector normal to the surface x^3+y^3+3xyz=3 ay the point(1,2,-1)?

1 Answer
Cesareo R.
Oct 25, 2016

{-1/sqrt[14], 3/sqrt[14], sqrt[2/7]}

Explanation:

Calling

f(x,y,z)=x^3+y^3+3xyz-3=0

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

The gradient is obtained as follows

grad f(x,y,z) = (f_x,f_y,f_z) = 3(x^2+yz,y^2+xz,xy) at point
(1,2,-1) has the value
3(-1,3,2) and the unit vector is
({-1,3,2})/sqrt(1+3^2+2^2)={-1/sqrt[14], 3/sqrt[14], sqrt[2/7]}

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