Question

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

Answer 1

`{-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]}`